A solution contains 0.2 M NH_4OH and 0.2 M | vedclass

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(A) The initial concentrations are $[Base] = 0.2 \ M$ and $[Salt] = 0.2 \ M$. When $1.0 \ mL$ of $0.001 \ M \ HCl$ is added to a large volume (assumin

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$2 	imes 10^{-5}$

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(A) The initial concentrations are $[Base] = 0.2 \ M$ and $[Salt] = 0.2 \ M$. When $1.0 \ mL$ of $0.001 \ M \ HCl$ is added to a large volume (assuming total volume remains approximately constant at $V$),the amount of $HCl$ added is negligible compared to the moles of $NH_4OH$ and $NH_4Cl$ present. Since $NH_4OH$ is a weak base and $NH_4Cl$ is its salt,the solution acts as a basic buffer. The Henderson-Hasselbalch equation for a basic buffer is: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$. Given $K_b = 2 	imes 10^{-5}$,$pK_b = -\log(2 	imes 10^{-5}) = 5 - \log 2 = 5 - 0.301 = 4.699$. Since the amount of $HCl$ added is extremely small $(10^{-6} \ mol)$,the change in concentrations of $NH_4OH$ and $NH_4Cl$ is negligible. Thus,$[OH^-] = K_b 	imes \frac{[Base]}{[Salt]} = 2 	imes 10^{-5} 	imes \frac{0.2}{0.2} = 2 	imes 10^{-5} \ M$.

Buffer Solution	Volume (in $mL$) of $0.1 \ M$ Weak acid	Volume (in $mL$) of $0.1 \ M$ sodium salt of weak acid
$I$	$4.0$	$4.0$
$II$	$4.0$	$40.0$
$III$	$40.0$	$4.0$
$IV$	$0.1$	$10.0$

$A$ solution contains $0.2 \ M \ NH_4OH$ and $0.2 \ M \ NH_4Cl$. If $1.0 \ mL$ of $0.001 \ M \ HCl$ is added,what will be the $[OH^-]$ of the resulting solution? [Given: $K_b = 2 \times 10^{-5}$]

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