A buffer solution of pH = 9.80 is to be prepa | vedclass

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(D) Given $pH = 9.80$,so $pOH = 14 - 9.80 = 4.20$. Using the Henderson-Hasselbalch equation for a basic buffer: $pOH = pK_b + \log \frac{[NH_4^+]}{[NH

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$218$

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(D) Given $pH = 9.80$,so $pOH = 14 - 9.80 = 4.20$. Using the Henderson-Hasselbalch equation for a basic buffer: $pOH = pK_b + \log \frac{[NH_4^+]}{[NH_3]}$ $4.20 = 4.74 + \log \frac{[NH_4^+]}{[NH_3]}$ $-0.54 = \log \frac{[NH_4^+]}{[NH_3]}$ Given $\log 0.28 = -0.54$,so $\frac{[NH_4^+]}{[NH_3]} = 0.28$. Let $V$ be the volume of $1 \ M \ NH_4Cl$ in $1000 \ mL$ of solution. Then the volume of $1 \ M \ NH_3$ is $(1000 - V) \ mL$. Since molarities are equal,the ratio of concentrations is equal to the ratio of volumes: $\frac{V}{1000 - V} = 0.28$ $V = 280 - 0.28V$ $1.28V = 280$ $V = \frac{280}{1.28} \approx 218.75 \ mL \approx 218 \ mL$.

$A$ buffer solution of $pH = 9.80$ is to be prepared from $1 \ M \ NH_4Cl$ and $1 \ M \ NH_3$ solution. $pK_{b}$ of $NH_3$ is $4.74$. In $1 \ L$ of the buffer solution,the volume of $NH_4Cl$ solution required is $(\log 0.28 = -0.54) :-$ (in $mL$)

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