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InterviewSolution
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Calculate the pH of a solution obtained by mixing 50 ml of 0.1 M NaOH with 100 ml of 0.1 M CH3COOH, Ka for CH3COOH = 1.8 x 10-5 ? |
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Answer» 50 ml of 0 ∙ 1 M NaOH contains = 5 m mol NaOH 100 ml of 0∙1 M CH3COOH contains = l0 m mol CH3COOH CH3COOH + NaOH → CH3COONa + H2O CH3COOH left unreacted = 10 – 5 = 5 m mol CH3COONa formed = 5 m mol Volume of solution = 150 [CH3COOH] = \(\frac{5}{150}\) = \(\frac{1}{30}\)M [CH3COONa] = \(\frac{5}{150}\) = \(\frac{1}{30}\) M This solution will behave as an acidic buffer. pH = pKa + log\(\frac{[Salt]}{[Acid]}\) = − log (1∙8 × 10−5) + log\(\frac{\frac{1}{30}}{\frac{1}{30}}\) = 4∙74. |
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