$3 g$ of acetic acid is added to $250 m L$ of $0.1 M H C l$ and the solution made up to $500 m L$ . To $20 m L$ of this solution $1 / 2 m L$ of $5 M N a O H$ is added. The $p H$ of the solution is

[Given: ${p K}_{a}$ of acetic acid $= 4.75$ , molar mass of acetic acid $= 60 g / m o l, l o g ? 3 = 0.4771$ ] Neglect any changes in volume.

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alok kumar singh | Contributor-Level 10

a month ago

$500 m L$ has $H C l = 25 \times 10^{- 3} m o l = 25 m \cdot m o l$

$120 m L$ has $= 1 m \cdot m o l$

and $500 m L$ has ${C H}_{3} C O O H = \frac{1}{20} \times 10^{3} m \cdot m o l$

so $20 m L$ has $\frac{10^{3}}{20} \times \frac{20}{500} = 2 m \cdot m o l$

$N a O H$ added in $20 m L$ is $5 \times \frac{1}{2} = 2.5 m \cdot m o l$

So, $N a O H$ (left) $+ {C H}_{3} C O O H ? {C H}_{3} C O O H + H_{2} O$
1.5
2
1.5

$\Rightarrow p H = p^{K_{a}} + l o g ? \frac{salt}{acid} = 4.75 + l o g ? 0.4771 = 5.2271$

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