Calculate the pH of a 0.5 M aqueous solution | vedclass

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(A) For a salt of a weak acid and a strong base,the $pH$ is given by the formula: $pH = 7 + \frac{1}{2}(pK_{a} + \log C)$.Given $pK_{b} = 4.70$,we cal

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$11$

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(A) For a salt of a weak acid and a strong base,the $pH$ is given by the formula: $pH = 7 + \frac{1}{2}(pK_{a} + \log C)$.Given $pK_{b} = 4.70$,we calculate $pK_{a} = 14 - 4.70 = 9.30$.Concentration $C = 0.5 \ M$.Substituting the values: $pH = 7 + \frac{1}{2}(9.30 + \log 0.5)$.Since $\log 0.5 = -0.30$,we have $pH = 7 + \frac{1}{2}(9.30 - 0.30) = 7 + \frac{1}{2}(9.0) = 7 + 4.5 = 11.5$.

Calculate the $pH$ of a $0.5 \ M$ aqueous solution of $KCN$. Given that $pK_{b}$ for $CN^{-} = 4.70$. (in $.5$)

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