Which of the following mixtures gives a buffe | vedclass

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(B) For a basic buffer, $pOH = pK_{b} + \log \frac{[Salt]}{[Base]}$. Given $pH = 9.25$, so $pOH = 14 - 9.25 = 4.75$. Substituting values: $4.75 = 4.75

Vedclass · Accepted Answer

$0.2 \ M \ NH_{4}OH \ (0.5 \ L) + 0.1 \ M \ HCl \ (0.5 \ L)$

Vedclass · Answer

(B) For a basic buffer, $pOH = pK_{b} + \log \frac{[Salt]}{[Base]}$. Given $pH = 9.25$, so $pOH = 14 - 9.25 = 4.75$. Substituting values: $4.75 = 4.75 + \log \frac{[Salt]}{[Base]}$, which implies $\log \frac{[Salt]}{[Base]} = 0$, so $[Salt] = [Base]$. This means the number of millimoles of $NH_{4}Cl$ (salt) formed must equal the remaining millimoles of $NH_{4}OH$ (base). Checking Option $(B)$: $NH_{4}OH + HCl ightarrow NH_{4}Cl + H_{2}O$ Initial millimoles of $NH_{4}OH = 0.2 \ M 	imes 500 \ mL = 100 \ mmol$. Initial millimoles of $HCl = 0.1 \ M 	imes 500 \ mL = 50 \ mmol$. After reaction, $50 \ mmol$ of $NH_{4}Cl$ is formed and $50 \ mmol$ of $NH_{4}OH$ remains. Since $[Salt] = [Base]$, the $pH$ will be $9.25$.

Which of the following mixtures gives a buffer solution with $pH = 9.25$?
Given: $pK_{b} (NH_{4}OH) = 4.75$

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Which of the following mixtures gives a buffer solution with $pH = 9.25$? Given: $pK_{b} (NH_{4}OH) = 4.75$