A buffer solution with pH = 9 is to be prepar | vedclass

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(D) For a basic buffer,the relationship is given by the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.Given $pH = 9$,we ca

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$1.8$

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(D) For a basic buffer,the relationship is given by the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.Given $pH = 9$,we calculate $pOH = 14 - 9 = 5$.Given $K_b = 1.8 	imes 10^{-5}$,we calculate $pK_b = -\log(1.8 	imes 10^{-5}) \approx 4.74$.Substituting the values into the equation: $5 = 4.74 + \log \frac{[NH_4Cl]}{1.0}$.Rearranging gives $\log [NH_4Cl] = 5 - 4.74 = 0.26$.Taking the antilog: $[NH_4Cl] = 10^{0.26} \approx 1.8 \ M$.Since the volume is $1 \ L$,the number of moles of $NH_4Cl$ required is $1.8 \ mol 	imes 1 \ L = 1.8 \ mol$.

$A$ buffer solution with $pH = 9$ is to be prepared by mixing $NH_4Cl$ and $NH_4OH$. Calculate the number of moles of $NH_4Cl$ that should be added to one litre of $1.0 \ M \ NH_4OH$. $[K_b = 1.8 \times 10^{-5}]$

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