Calculate the pH on mixing 100 mL of 0.2 M | vedclass

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(B) The reaction is: $HCN + KOH 	o KCN + H_2O$Initial moles: $n(HCN) = 0.1 \ L 	imes 0.2 \ M = 0.02 \ mol$,$n(KOH) = 0.4 \ L 	imes 0.05 \ M = 0.02

Vedclass · Accepted Answer

$9.3$

Vedclass · Answer

(B) The reaction is: $HCN + KOH 	o KCN + H_2O$Initial moles: $n(HCN) = 0.1 \ L 	imes 0.2 \ M = 0.02 \ mol$,$n(KOH) = 0.4 \ L 	imes 0.05 \ M = 0.02 \ mol$.Since the moles are equal,they react completely to form $0.02 \ mol$ of $KCN$.Total volume = $100 \ mL + 400 \ mL = 500 \ mL = 0.5 \ L$.Concentration of salt $[KCN] = \frac{0.02 \ mol}{0.5 \ L} = 0.04 \ M = 4 	imes 10^{-2} \ M$.Given $pK_b(CN^-) = 8$,so $pK_a(HCN) = 14 - 8 = 6$.For a salt of a weak acid and strong base,$pH = 7 + \frac{1}{2}(pK_a + \log C)$.$pH = 7 + \frac{1}{2}(6 + \log(4 	imes 10^{-2}))$$pH = 7 + \frac{1}{2}(6 + \log 4 - 2) = 7 + \frac{1}{2}(4 + 0.6) = 7 + 2.3 = 9.3$.

Calculate the $pH$ on mixing $100 \ mL$ of $0.2 \ M \ HCN$ with $400 \ mL$ of $0.05 \ M \ KOH$ at $25 \ ^\circ C$ ($pK_b$ for $CN^- = 8$).

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