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6,301 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=0.8613 mol, V=47.11 L, T=489.2 K | For an ideal gas, P V = n R T. With n = 0.8613 mol, V = 47.11 L, T = 489.2 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.7339 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,302 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=3.037 mol, V=4.138 L, T=402.3 K | For an ideal gas, P V = n R T. With n = 3.037 mol, V = 4.138 L, T = 402.3 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 24.23 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,303 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.262 mol, V=21.2 L, T=334.1 K | For an ideal gas, P V = n R T. With n = 2.262 mol, V = 21.2 L, T = 334.1 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 2.924 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,304 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=0.4702 mol, V=34.92 L, T=573.5 K | For an ideal gas, P V = n R T. With n = 0.4702 mol, V = 34.92 L, T = 573.5 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.6336 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,305 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=4.288 mol, V=26.22 L, T=317.6 K | For an ideal gas, P V = n R T. With n = 4.288 mol, V = 26.22 L, T = 317.6 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 4.261 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,306 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.042 mol, V=41.56 L, T=498.9 K | For an ideal gas, P V = n R T. With n = 2.042 mol, V = 41.56 L, T = 498.9 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 2.011 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,307 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=3.156 mol, V=17.21 L, T=511.2 K | For an ideal gas, P V = n R T. With n = 3.156 mol, V = 17.21 L, T = 511.2 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 7.695 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,308 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.467 mol, V=22.05 L, T=358.2 K | For an ideal gas, P V = n R T. With n = 2.467 mol, V = 22.05 L, T = 358.2 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 3.288 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,309 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=3.956 mol, V=34.84 L, T=218.5 K | For an ideal gas, P V = n R T. With n = 3.956 mol, V = 34.84 L, T = 218.5 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 2.036 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,310 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=0.5914 mol, V=39.13 L, T=251.5 K | For an ideal gas, P V = n R T. With n = 0.5914 mol, V = 39.13 L, T = 251.5 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.312 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,311 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=0.09966 mol, V=10.05 L, T=216.9 K | For an ideal gas, P V = n R T. With n = 0.09966 mol, V = 10.05 L, T = 216.9 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.1765 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,312 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=4.951 mol, V=10.56 L, T=408.1 K | For an ideal gas, P V = n R T. With n = 4.951 mol, V = 10.56 L, T = 408.1 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 15.71 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,313 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=1.959 mol, V=4.875 L, T=553.8 K | For an ideal gas, P V = n R T. With n = 1.959 mol, V = 4.875 L, T = 553.8 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 18.26 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,314 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=4.357 mol, V=8.123 L, T=210.9 K | For an ideal gas, P V = n R T. With n = 4.357 mol, V = 8.123 L, T = 210.9 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 9.284 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,315 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.284 mol, V=37.39 L, T=328.2 K | For an ideal gas, P V = n R T. With n = 2.284 mol, V = 37.39 L, T = 328.2 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.645 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,316 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=0.5087 mol, V=11.32 L, T=270.2 K | For an ideal gas, P V = n R T. With n = 0.5087 mol, V = 11.32 L, T = 270.2 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.9964 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,317 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=4.488 mol, V=40.96 L, T=342.3 K | For an ideal gas, P V = n R T. With n = 4.488 mol, V = 40.96 L, T = 342.3 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 3.078 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,318 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.55 mol, V=28.82 L, T=273.4 K | For an ideal gas, P V = n R T. With n = 2.55 mol, V = 28.82 L, T = 273.4 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.985 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,319 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.432 mol, V=21.51 L, T=539 K | For an ideal gas, P V = n R T. With n = 2.432 mol, V = 21.51 L, T = 539 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 5 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,320 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.811 mol, V=31.25 L, T=561 K | For an ideal gas, P V = n R T. With n = 2.811 mol, V = 31.25 L, T = 561 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 4.141 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,321 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=4.92 mol, V=14.87 L, T=532.6 K | For an ideal gas, P V = n R T. With n = 4.92 mol, V = 14.87 L, T = 532.6 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 14.45 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,322 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=4.816 mol, V=15.55 L, T=246.3 K | For an ideal gas, P V = n R T. With n = 4.816 mol, V = 15.55 L, T = 246.3 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 6.261 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,323 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.777 mol, V=49.26 L, T=517.5 K | For an ideal gas, P V = n R T. With n = 2.777 mol, V = 49.26 L, T = 517.5 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 2.394 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,324 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=3.903 mol, V=12.25 L, T=337 K | For an ideal gas, P V = n R T. With n = 3.903 mol, V = 12.25 L, T = 337 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 8.811 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,325 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=4.316 mol, V=36.96 L, T=565.1 K | For an ideal gas, P V = n R T. With n = 4.316 mol, V = 36.96 L, T = 565.1 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 5.414 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,326 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.648 mol, V=10.99 L, T=363 K | For an ideal gas, P V = n R T. With n = 2.648 mol, V = 10.99 L, T = 363 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 7.175 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,327 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=3.229 mol, V=6.645 L, T=445.8 K | For an ideal gas, P V = n R T. With n = 3.229 mol, V = 6.645 L, T = 445.8 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 17.78 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,328 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=0.5213 mol, V=34.71 L, T=348.2 K | For an ideal gas, P V = n R T. With n = 0.5213 mol, V = 34.71 L, T = 348.2 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.4292 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,329 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=0.1001 mol, V=25.62 L, T=528.9 K | For an ideal gas, P V = n R T. With n = 0.1001 mol, V = 25.62 L, T = 528.9 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.1696 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,330 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=4.966 mol, V=3.71 L, T=432.3 K | For an ideal gas, P V = n R T. With n = 4.966 mol, V = 3.71 L, T = 432.3 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 47.49 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,331 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=1.454 mol, V=46.78 L, T=429.6 K | For an ideal gas, P V = n R T. With n = 1.454 mol, V = 46.78 L, T = 429.6 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.096 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,332 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=3.576 mol, V=13.13 L, T=381.2 K | For an ideal gas, P V = n R T. With n = 3.576 mol, V = 13.13 L, T = 381.2 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 8.518 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,333 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=4.039 mol, V=30.09 L, T=361.1 K | For an ideal gas, P V = n R T. With n = 4.039 mol, V = 30.09 L, T = 361.1 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 3.977 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,334 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.868 mol, V=21.33 L, T=577 K | For an ideal gas, P V = n R T. With n = 2.868 mol, V = 21.33 L, T = 577 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 6.368 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,335 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=3.126 mol, V=43.54 L, T=523.2 K | For an ideal gas, P V = n R T. With n = 3.126 mol, V = 43.54 L, T = 523.2 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 3.082 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,336 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=3.287 mol, V=32.27 L, T=532.1 K | For an ideal gas, P V = n R T. With n = 3.287 mol, V = 32.27 L, T = 532.1 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 4.448 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
6,337 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.01899 M | A strong monoprotic acid is fully dissociated. At concentration 0.01899 mol/L, [H⁺] = 0.01899 M and pH = −log₁₀[H⁺] = 1.722. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,338 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.0173 M | A strong monoprotic acid is fully dissociated. At concentration 0.0173 mol/L, [H⁺] = 0.0173 M and pH = −log₁₀[H⁺] = 1.762. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,339 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.01066 M | A strong monoprotic acid is fully dissociated. At concentration 0.01066 mol/L, [H⁺] = 0.01066 M and pH = −log₁₀[H⁺] = 1.972. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,340 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.0192 M | A strong monoprotic acid is fully dissociated. At concentration 0.0192 mol/L, [H⁺] = 0.0192 M and pH = −log₁₀[H⁺] = 1.717. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,341 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.03653 M | A strong monoprotic acid is fully dissociated. At concentration 0.03653 mol/L, [H⁺] = 0.03653 M and pH = −log₁₀[H⁺] = 1.437. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,342 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 1.8577e-04 M | A strong monoprotic acid is fully dissociated. At concentration 1.8577e-04 mol/L, [H⁺] = 1.8577e-04 M and pH = −log₁₀[H⁺] = 3.731. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,343 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.003687 M | A strong monoprotic acid is fully dissociated. At concentration 0.003687 mol/L, [H⁺] = 0.003687 M and pH = −log₁₀[H⁺] = 2.433. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,344 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 2.7261e-04 M | A strong monoprotic acid is fully dissociated. At concentration 2.7261e-04 mol/L, [H⁺] = 2.7261e-04 M and pH = −log₁₀[H⁺] = 3.564. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,345 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.02242 M | A strong monoprotic acid is fully dissociated. At concentration 0.02242 mol/L, [H⁺] = 0.02242 M and pH = −log₁₀[H⁺] = 1.649. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,346 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.005468 M | A strong monoprotic acid is fully dissociated. At concentration 0.005468 mol/L, [H⁺] = 0.005468 M and pH = −log₁₀[H⁺] = 2.262. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,347 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.008135 M | A strong monoprotic acid is fully dissociated. At concentration 0.008135 mol/L, [H⁺] = 0.008135 M and pH = −log₁₀[H⁺] = 2.09. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,348 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.005085 M | A strong monoprotic acid is fully dissociated. At concentration 0.005085 mol/L, [H⁺] = 0.005085 M and pH = −log₁₀[H⁺] = 2.294. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,349 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 2.7593e-04 M | A strong monoprotic acid is fully dissociated. At concentration 2.7593e-04 mol/L, [H⁺] = 2.7593e-04 M and pH = −log₁₀[H⁺] = 3.559. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,350 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.004966 M | A strong monoprotic acid is fully dissociated. At concentration 0.004966 mol/L, [H⁺] = 0.004966 M and pH = −log₁₀[H⁺] = 2.304. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,351 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 7.6423e-04 M | A strong monoprotic acid is fully dissociated. At concentration 7.6423e-04 mol/L, [H⁺] = 7.6423e-04 M and pH = −log₁₀[H⁺] = 3.117. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,352 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.02074 M | A strong monoprotic acid is fully dissociated. At concentration 0.02074 mol/L, [H⁺] = 0.02074 M and pH = −log₁₀[H⁺] = 1.683. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,353 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.04592 M | A strong monoprotic acid is fully dissociated. At concentration 0.04592 mol/L, [H⁺] = 0.04592 M and pH = −log₁₀[H⁺] = 1.338. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,354 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.00511 M | A strong monoprotic acid is fully dissociated. At concentration 0.00511 mol/L, [H⁺] = 0.00511 M and pH = −log₁₀[H⁺] = 2.292. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,355 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.02382 M | A strong monoprotic acid is fully dissociated. At concentration 0.02382 mol/L, [H⁺] = 0.02382 M and pH = −log₁₀[H⁺] = 1.623. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,356 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.0372 M | A strong monoprotic acid is fully dissociated. At concentration 0.0372 mol/L, [H⁺] = 0.0372 M and pH = −log₁₀[H⁺] = 1.429. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,357 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.07242 M | A strong monoprotic acid is fully dissociated. At concentration 0.07242 mol/L, [H⁺] = 0.07242 M and pH = −log₁₀[H⁺] = 1.14. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,358 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.04695 M | A strong monoprotic acid is fully dissociated. At concentration 0.04695 mol/L, [H⁺] = 0.04695 M and pH = −log₁₀[H⁺] = 1.328. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,359 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 2.1138e-04 M | A strong monoprotic acid is fully dissociated. At concentration 2.1138e-04 mol/L, [H⁺] = 2.1138e-04 M and pH = −log₁₀[H⁺] = 3.675. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,360 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.03293 M | A strong monoprotic acid is fully dissociated. At concentration 0.03293 mol/L, [H⁺] = 0.03293 M and pH = −log₁₀[H⁺] = 1.482. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,361 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.02638 M | A strong monoprotic acid is fully dissociated. At concentration 0.02638 mol/L, [H⁺] = 0.02638 M and pH = −log₁₀[H⁺] = 1.579. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,362 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 4.6675e-04 M | A strong monoprotic acid is fully dissociated. At concentration 4.6675e-04 mol/L, [H⁺] = 4.6675e-04 M and pH = −log₁₀[H⁺] = 3.331. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,363 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.02196 M | A strong monoprotic acid is fully dissociated. At concentration 0.02196 mol/L, [H⁺] = 0.02196 M and pH = −log₁₀[H⁺] = 1.658. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,364 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.02132 M | A strong monoprotic acid is fully dissociated. At concentration 0.02132 mol/L, [H⁺] = 0.02132 M and pH = −log₁₀[H⁺] = 1.671. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,365 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.001948 M | A strong monoprotic acid is fully dissociated. At concentration 0.001948 mol/L, [H⁺] = 0.001948 M and pH = −log₁₀[H⁺] = 2.71. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,366 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.004918 M | A strong monoprotic acid is fully dissociated. At concentration 0.004918 mol/L, [H⁺] = 0.004918 M and pH = −log₁₀[H⁺] = 2.308. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,367 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.07619 M | A strong monoprotic acid is fully dissociated. At concentration 0.07619 mol/L, [H⁺] = 0.07619 M and pH = −log₁₀[H⁺] = 1.118. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,368 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 1.4599e-04 M | A strong monoprotic acid is fully dissociated. At concentration 1.4599e-04 mol/L, [H⁺] = 1.4599e-04 M and pH = −log₁₀[H⁺] = 3.836. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,369 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.007356 M | A strong monoprotic acid is fully dissociated. At concentration 0.007356 mol/L, [H⁺] = 0.007356 M and pH = −log₁₀[H⁺] = 2.133. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,370 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 1.1719e-04 M | A strong monoprotic acid is fully dissociated. At concentration 1.1719e-04 mol/L, [H⁺] = 1.1719e-04 M and pH = −log₁₀[H⁺] = 3.931. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,371 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.004718 M | A strong monoprotic acid is fully dissociated. At concentration 0.004718 mol/L, [H⁺] = 0.004718 M and pH = −log₁₀[H⁺] = 2.326. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,372 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 2.9174e-04 M | A strong monoprotic acid is fully dissociated. At concentration 2.9174e-04 mol/L, [H⁺] = 2.9174e-04 M and pH = −log₁₀[H⁺] = 3.535. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,373 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.008331 M | A strong monoprotic acid is fully dissociated. At concentration 0.008331 mol/L, [H⁺] = 0.008331 M and pH = −log₁₀[H⁺] = 2.079. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,374 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.002353 M | A strong monoprotic acid is fully dissociated. At concentration 0.002353 mol/L, [H⁺] = 0.002353 M and pH = −log₁₀[H⁺] = 2.628. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,375 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.001791 M | A strong monoprotic acid is fully dissociated. At concentration 0.001791 mol/L, [H⁺] = 0.001791 M and pH = −log₁₀[H⁺] = 2.747. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,376 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.0473 M | A strong monoprotic acid is fully dissociated. At concentration 0.0473 mol/L, [H⁺] = 0.0473 M and pH = −log₁₀[H⁺] = 1.325. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,377 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.00843 M | A strong monoprotic acid is fully dissociated. At concentration 0.00843 mol/L, [H⁺] = 0.00843 M and pH = −log₁₀[H⁺] = 2.074. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,378 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 3.3246e-04 M | A strong monoprotic acid is fully dissociated. At concentration 3.3246e-04 mol/L, [H⁺] = 3.3246e-04 M and pH = −log₁₀[H⁺] = 3.478. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,379 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 4.4435e-04 M | A strong monoprotic acid is fully dissociated. At concentration 4.4435e-04 mol/L, [H⁺] = 4.4435e-04 M and pH = −log₁₀[H⁺] = 3.352. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,380 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.004841 M | A strong monoprotic acid is fully dissociated. At concentration 0.004841 mol/L, [H⁺] = 0.004841 M and pH = −log₁₀[H⁺] = 2.315. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,381 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.0964 M | A strong monoprotic acid is fully dissociated. At concentration 0.0964 mol/L, [H⁺] = 0.0964 M and pH = −log₁₀[H⁺] = 1.016. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,382 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 4.8343e-04 M | A strong monoprotic acid is fully dissociated. At concentration 4.8343e-04 mol/L, [H⁺] = 4.8343e-04 M and pH = −log₁₀[H⁺] = 3.316. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,383 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.00661 M | A strong monoprotic acid is fully dissociated. At concentration 0.00661 mol/L, [H⁺] = 0.00661 M and pH = −log₁₀[H⁺] = 2.18. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,384 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 9.4915e-04 M | A strong monoprotic acid is fully dissociated. At concentration 9.4915e-04 mol/L, [H⁺] = 9.4915e-04 M and pH = −log₁₀[H⁺] = 3.023. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,385 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.009719 M | A strong monoprotic acid is fully dissociated. At concentration 0.009719 mol/L, [H⁺] = 0.009719 M and pH = −log₁₀[H⁺] = 2.012. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,386 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.05292 M | A strong monoprotic acid is fully dissociated. At concentration 0.05292 mol/L, [H⁺] = 0.05292 M and pH = −log₁₀[H⁺] = 1.276. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,387 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.008924 M | A strong monoprotic acid is fully dissociated. At concentration 0.008924 mol/L, [H⁺] = 0.008924 M and pH = −log₁₀[H⁺] = 2.049. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,388 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.04961 M | A strong monoprotic acid is fully dissociated. At concentration 0.04961 mol/L, [H⁺] = 0.04961 M and pH = −log₁₀[H⁺] = 1.304. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,389 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.003636 M | A strong monoprotic acid is fully dissociated. At concentration 0.003636 mol/L, [H⁺] = 0.003636 M and pH = −log₁₀[H⁺] = 2.439. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,390 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.009371 M | A strong monoprotic acid is fully dissociated. At concentration 0.009371 mol/L, [H⁺] = 0.009371 M and pH = −log₁₀[H⁺] = 2.028. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,391 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.004825 M | A strong monoprotic acid is fully dissociated. At concentration 0.004825 mol/L, [H⁺] = 0.004825 M and pH = −log₁₀[H⁺] = 2.317. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,392 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 1.1285e-04 M | A strong monoprotic acid is fully dissociated. At concentration 1.1285e-04 mol/L, [H⁺] = 1.1285e-04 M and pH = −log₁₀[H⁺] = 3.948. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,393 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.0286 M | A strong monoprotic acid is fully dissociated. At concentration 0.0286 mol/L, [H⁺] = 0.0286 M and pH = −log₁₀[H⁺] = 1.544. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,394 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 7.1467e-04 M | A strong monoprotic acid is fully dissociated. At concentration 7.1467e-04 mol/L, [H⁺] = 7.1467e-04 M and pH = −log₁₀[H⁺] = 3.146. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,395 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 1.0214e-04 M | A strong monoprotic acid is fully dissociated. At concentration 1.0214e-04 mol/L, [H⁺] = 1.0214e-04 M and pH = −log₁₀[H⁺] = 3.991. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,396 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.001082 M | A strong monoprotic acid is fully dissociated. At concentration 0.001082 mol/L, [H⁺] = 0.001082 M and pH = −log₁₀[H⁺] = 2.966. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,397 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.001669 M | A strong monoprotic acid is fully dissociated. At concentration 0.001669 mol/L, [H⁺] = 0.001669 M and pH = −log₁₀[H⁺] = 2.778. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,398 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.009789 M | A strong monoprotic acid is fully dissociated. At concentration 0.009789 mol/L, [H⁺] = 0.009789 M and pH = −log₁₀[H⁺] = 2.009. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,399 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.08806 M | A strong monoprotic acid is fully dissociated. At concentration 0.08806 mol/L, [H⁺] = 0.08806 M and pH = −log₁₀[H⁺] = 1.055. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
6,400 | chemistry | acids_bases | strong_acid_ph | 4 | worked_example | pH of strong acid at concentration 0.0509 M | A strong monoprotic acid is fully dissociated. At concentration 0.0509 mol/L, [H⁺] = 0.0509 M and pH = −log₁₀[H⁺] = 1.293. This relation follows directly from the definition of pH and the complete dissociation assumption. | pH = -log10 [H+] | mole_concept; logarithmic functions | Calculate the pH of a strong monoprotic acid solution. |
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