20 mL of 0.1 M NaOH is added to 50 mL of 0.1 M acetic acid solution. The pH of the resulting solution is ________ (Nearest integer) Given: = 4.76 log 2 = 0.30 log 3 = 0.48 [2023]

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Solution

Q.
20 mL of 0.1 M NaOH is added to 50 mL of 0.1 M acetic acid solution. The pH of the resulting solution is ________ $\times 10^{- 2}$ (Nearest integer)

Given: ${pK}_{a} ({CH}_{3} COOH)$ = 4.76

log 2 = 0.30

log 3 = 0.48 [2023]

Ans.
(458)

   $NaOH$    $+$     ${CH}_{3} COOH \to {CH}_{3} COONa + H_{2} O$

   $0.1 M, 20 ml 0.1 M, 50 ml$

$Millimole = 0.1 \times 20 = 2, 0.1 \times 50 = 5$

   $L.R. = NaOH$

$So$ $0$ $5 - 2 = (3)$ $(2)$

$So resultant solution is an acidic buffer solution$

So $pH = p K_{a} + \log (\frac{salt}{acid})$

$pH = 4.76 + \log \frac{2}{3}$

$pH = 4.76 + (- 0.18) = 4.58 = 458 \times 10^{- 2}$

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