If a very small amount of HPh (phenolphthalei | vedclass

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(B) For the salt of a weak acid and strong base,the $pH$ is calculated as: $pH = \frac{1}{2}(pK_w + pK_a(CH_3COOH) + \log C)$ Given $K_b(CH_3COO^-) =

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$0.8$

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(B) For the salt of a weak acid and strong base,the $pH$ is calculated as: $pH = \frac{1}{2}(pK_w + pK_a(CH_3COOH) + \log C)$ Given $K_b(CH_3COO^-) = 10^{-9}$,so $K_a(CH_3COOH) = \frac{K_w}{K_b} = \frac{10^{-14}}{10^{-9}} = 10^{-5}$,which means $pK_a = 5$. $pH = \frac{1}{2}(14 + 5 + \log 0.1) = \frac{1}{2}(19 - 1) = 9$. For the indicator $HPh ightleftharpoons H^+ + Ph^-$,the Henderson-Hasselbalch equation is: $pH = pK_a(HPh) + \log \frac{[Ph^-]}{[HPh]}$ $9 = 9.6 + \log \frac{[Ph^-]}{[HPh]}$ $\log \frac{[Ph^-]}{[HPh]} = -0.6 = -2 	imes 0.3 = -2 \log 2 = \log(2^{-2}) = \log(0.25)$. So,$\frac{[Ph^-]}{[HPh]} = 0.25 = \frac{1}{4}$. Let the total concentration be $[HPh] + [Ph^-] = 1$. Since $[HPh] = 4[Ph^-]$,we have $4[Ph^-] + [Ph^-] = 1$,so $5[Ph^-] = 1$,$[Ph^-] = 0.2$. Then $[HPh] = 1 - 0.2 = 0.8$. The colourless form is $HPh$,so the fraction is $0.8$.

If a very small amount of $HPh$ (phenolphthalein) is added to a $0.1 \ M$ aqueous solution of $CH_3COONa$,the fraction of the indicator that exists in the colourless form is:
$[K_b(CH_3COO^{-}) = 10^{-9}, pK_a(HPh) = 9.6, \log 2 = 0.3]$

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If a very small amount of $HPh$ (phenolphthalein) is added to a $0.1 \ M$ aqueous solution of $CH_3COONa$,the fraction of the indicator that exists in the colourless form is: $[K_b(CH_3COO^{-}) = 10^{-9}, pK_a(HPh) = 9.6, \log 2 = 0.3]$