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Calculate the pH of a solution which is 0.1 M in HA and 0.5 M in NaA. Ka for HA is 1.8 × 10-6 ? |
Answer» HA + H2O ⇌ H3O+ + A− NaA → Na+ + A- pH = pKa + log\(\frac{[Salt]}{[acid]}\) = − log (1∙8 × 10−6) + log \(\frac{[0.5]}{[0.1]}\) = 5∙744 + 0∙6990 = 6∙44. |
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