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(B) For a basic buffer,the $pOH$ is calculated using the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$ Given: $[Base] = [N

Vedclass · Accepted Answer

$9.43$

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(B) For a basic buffer,the $pOH$ is calculated using the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$ Given: $[Base] = [NH_3] = 0.30 \ M$,$[Salt] = [NH_4^+] = 0.20 \ M$,and $K_b = 1.8 	imes 10^{-5}$. $pK_b = -\log(1.8 	imes 10^{-5}) = 5 - \log(1.8) = 5 - 0.255 = 4.745$. $pOH = 4.745 + \log \frac{0.20}{0.30} = 4.745 + \log(0.667) = 4.745 - 0.176 = 4.569$. Since $pH + pOH = 14$,$pH = 14 - 4.569 = 9.431 \approx 9.43$.

$A$ buffer solution is prepared in which the concentration of $NH_3$ is $0.30 \ M$ and the concentration of $NH_4^+$ is $0.20 \ M.$ If the equilibrium constant,$K_b$ for $NH_3$ equals $1.8 \times 10^{-5},$ what is the $pH$ of this solution? $(\log \ 2.7 = 0.43)$

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