100 mL of a solution contains 0.1 M NH_4OH | vedclass

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(D) The given solution contains a weak base $(NH_4OH)$ and its salt with a strong acid $(NH_4Cl)$,which forms a basic buffer solution.$pH$ of a buffer

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$10 \ mL$ of distilled water

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(D) The given solution contains a weak base $(NH_4OH)$ and its salt with a strong acid $(NH_4Cl)$,which forms a basic buffer solution.$pH$ of a buffer solution is given by the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.Adding a small amount of distilled water to a buffer solution changes the concentration of both the salt and the base by the same dilution factor,so the ratio $\frac{[Salt]}{[Base]}$ remains constant.Therefore,the $pH$ (and $pOH$) of the buffer solution remains unchanged upon dilution with water.Adding $NH_4OH$,$NH_4Cl$,or $NaOH$ would change the ratio of salt to base,thereby altering the $pH$.

$100 \ mL$ of a solution contains $0.1 \ M$ $NH_4OH$ and $0.1 \ M$ $NH_4Cl$. The $pH$ of the solution will not change on adding :-

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