If volume ratio of $0.02 M$ $H C l$ and $0.02 M$ $B O H$ (a weak base) mixed to make $200$ mL of buffer of $p H = 7$ is $2.23 : X$ . The value of $X$ is:

$K_{b}$ for $B O H = 8.1 \times 10^{- 8}$ .

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Solution

The correct option is B 5
$p K_{b} = - l o g K_{b} = - l o g (8.1 \times 10^{- 8}) = 7.096$
$p H = 14 - p K_{b} - l o g \frac{[B C l]}{[B O H]}$
$7 = 14 - 7.096 - l o g \frac{2.23 \times 0.02}{(X - 2.23) \times 0.02}$
$l o g \frac{2.23}{X - 2.23} = - 0.096$

$\frac{2.23}{X - 2.23} = 0.813$ $⟹ X = 5$

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If volume ratio of 0.02M HCl and 0.02M BOH (a weak base) mixed to make 200 mL of buffer of pH=7 is 2.23:X. The value of X is:Kb for BOH=8.1×10−8.

The correct option is B 5pKb=−logKb=−log(8.1×10−8)=7.096pH=14−pKb−log[BCl][BOH]7=14−7.096−log2.23×0.02(X−2.23)×0.02log2.23X−2.23=−0.0962.23X−2.23=0.813 ⟹X=5

If volume ratio of $0.02 M$ $H C l$ and $0.02 M$ $B O H$ (a weak base) mixed to make $200$ mL of buffer of $p H = 7$ is $2.23 : X$ . The value of $X$ is:

$K_{b}$ for $B O H = 8.1 \times 10^{- 8}$ .