In a buffer solution,the ratio of NH_4Cl to N | vedclass

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(B) For a basic buffer,the $pOH$ is given by the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.Initially,the ratio $\frac{

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(B) For a basic buffer,the $pOH$ is given by the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.Initially,the ratio $\frac{[Salt]}{[Base]} = \frac{1}{1} = 1$,so $pOH = pK_b + \log(1) = pK_b$.Since $pH + pOH = 14$,the initial $pH = 14 - pK_b$.When the ratio becomes $\frac{2}{1} = 2$,the new $pOH' = pK_b + \log(2)$.Since $\log(2) \approx 0.301$,the $pOH$ increases.As $pH = 14 - pOH$,an increase in $pOH$ results in a decrease in the $pH$ value.

In a buffer solution,the ratio of $NH_4Cl$ to $NH_4OH$ is $1:1$. If this ratio is changed to $2:1$,the $pH$ value of the buffer will ...

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