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 $pH$.

Buffer Solution	Volume of $0.1 \ M$ Weak Acid (mL)	Volume of $0.1 \ M$ Sodium Salt (mL)
$I$	$4.0$	$4.0$
$II$	$4.0$	$40.0$
$III$	$40.0$	$4.0$
$IV$	$0.1$	$10.0$

$100 \ mL$ 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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Calculate the $pH$ of a buffer solution containing $0.01 \ M$ salt and $0.004 \ M$ weak acid. $(pK_{a} = 4.762)$

Buffer Solution Volume of $0.1 \ M$ Weak Acid (mL) Volume of $0.1 \ M$ Sodium Salt (mL)

$I$ $4.0$ $4.0$

$II$ $4.0$ $40.0$

$III$ $40.0$ $4.0$

$IV$ $0.1$ $10.0$

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