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(A) For a basic buffer,the formula is $pOH = pK_b + \log \left( \frac{[Salt]}{[Base]} \right)$.Given: $[Salt] = [NH_4^+] = 0.20 \ M$,$[Base] = [NH_3]

Vedclass · Accepted Answer

$9.43$

Vedclass · Answer

(A) For a basic buffer,the formula is $pOH = pK_b + \log \left( \frac{[Salt]}{[Base]} ight)$.Given: $[Salt] = [NH_4^+] = 0.20 \ M$,$[Base] = [NH_3] = 0.30 \ M$,and $K_b = 1.8 	imes 10^{-5}$.First,calculate $pK_b = -\log(K_b) = -\log(1.8 	imes 10^{-5}) \approx 4.74$.Now,$pOH = 4.74 + \log \left( \frac{0.20}{0.30} ight) = 4.74 + \log(0.667) = 4.74 - 0.176 = 4.564$.Finally,$pH = 14 - pOH = 14 - 4.564 = 9.436 \approx 9.43$.

$A$ buffer solution is prepared with a concentration of $0.30 \ M \ NH_3$ and $0.20 \ M \ NH_4^+$. If the equilibrium constant for $NH_3$ is $K_b = 1.8 \times 10^{-5}$,what will be the $pH$ of this solution?

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