A weak monobasic acid is titrated with aqueous KOH solution. What is the $p H$ at the point of one third neutralization of the acid ? ( $K_{a}$ is $10^{- 5}$ )

5 - log 2

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5 - log 3

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5 - log 4

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5 + log 3

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Solution

The correct option is A 5 - log 2
$K_{a} = 10^{- 5}$ and $p K_{a} = 5$
$H A + K O H \to K A + H_{2} O$
$\begin{matrix} i n i t i a l 1000 f i n a l 1 - \frac{1}{3} . \frac{1}{3} \frac{1}{3} \end{matrix}$ moles
$= \frac{2}{3}$ mole $\frac{1}{3} m o l e$
Final solution acts as a buffer
$p H = p K_{a} + l o g \frac{[S a l t]}{[A c i d]}$
$p H = 5 + l o g \frac{1 / 3}{2 / 3} = 5 + l o g \frac{1}{2} = 5 - l o g 2$

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A weak monobasic acid is titrated with aqueous KOH solution. What is the pH at the point of one third neutralization of the acid ? (Ka is 10−5)

The correct option is A 5 - log 2Ka=10−5 and pKa=5 HA+KOH→KA+H2O initial 1 0 0 0final 1−13 . 13 13moles =23 mole 13mole Final solution acts as a buffer pH=pKa+log[Salt][Acid] pH=5+log1/32/3=5+log12=5−log2

A weak monobasic acid is titrated with aqueous KOH solution. What is the $p H$ at the point of one third neutralization of the acid ? ( $K_{a}$ is $10^{- 5}$ )