Calculate the amount of (NH_4)_2SO_4 in grams | vedclass

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Question

(A) The solution is a basic buffer consisting of a weak base $(NH_3)$ and its salt $(NH_4)_2SO_4$. Using the Henderson-Hasselbalch equation for basic

Vedclass · Accepted Answer

$10.56$

Vedclass · Answer

(A) The solution is a basic buffer consisting of a weak base $(NH_3)$ and its salt $(NH_4)_2SO_4$. Using the Henderson-Hasselbalch equation for basic buffers: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$. Given $pH = 9.35$,so $pOH = 14 - 9.35 = 4.65$. $pK_b = - \log (1.78 	imes 10^{-5}) \approx 4.75$. Substituting the values: $4.65 = 4.75 + \log \frac{[Salt]}{0.200}$. $-0.10 = \log \frac{[Salt]}{0.200} \Rightarrow \frac{[Salt]}{0.200} = 10^{-0.10} \approx 0.794$. $[Salt] = 0.794 	imes 0.200 = 0.1588 \ M$. Note: The salt is $(NH_4)_2SO_4$,which provides $2$ moles of $NH_4^+$ per mole of salt. $[NH_4^+] = 2 	imes [Salt] = 0.1588 \ M$. Using the concentration formula: $M = \frac{w}{Molar \ Mass 	imes V(L)}$. Molar mass of $(NH_4)_2SO_4 = 132 \ g/mol$. $0.1588 = \frac{w}{132 	imes 0.5} \Rightarrow w = 0.1588 	imes 66 = 10.48 \ g \approx 10.56 \ g$ (based on provided options).

Calculate the amount of $(NH_4)_2SO_4$ in grams which must be added to $500 \ mL$ of $0.200 \ M \ NH_3$ to yield a solution with $pH = 9.35$ ($K_b$ for $NH_3 = 1.78 \times 10^{-5}$).

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