Based on acetic acid Ka∘=1.8×10−5 solution,If the solution obtained in Q.99 is diluted 10 times, it pH becomes

A
4.75
B
3.75
C
5.75
D
6.25

Solution:

$\begin{array}{l} M ({CH}_{3} COONa) = 82 g {mol}^{- 1} \\ n ({CII}_{3} {COONa}^{-} m / M - 0.82 g / 82 g {mol}^{- 1} - 0.01 mol \\ c ({CH}_{3} COONa) = n / V = 0.01 mol / 0.1 L = 0.1 M \\ pH = p K_{a} + \log [salt] / [acid] = - \log (1.8 \times 10^{- 5}) + \log (0.1 / 0.1) = 4.75 \end{array}$

from the given condition the solution is diluted 10 times. then

The pH of buffer solution does not change on dilution.

Based on acetic acid $(K_{a}^{\circ} = 1.8 \times 10^{- 5})$ solution,
If the solution obtained in Q.99 is diluted 10 times, it pH becomes

Solution:

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