Calculate pH of the solution obtained by mixi | vedclass

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(D) The reaction between $NH_4Cl$ and $NaOH$ is: $NH_4Cl + NaOH ightarrow NH_4OH + NaCl$ Initial moles: $n(NH_4Cl) = 0.2 \ M 	imes 0.050 \ L = 10 \

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$9.74$

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(D) The reaction between $NH_4Cl$ and $NaOH$ is: $NH_4Cl + NaOH ightarrow NH_4OH + NaCl$ Initial moles: $n(NH_4Cl) = 0.2 \ M 	imes 0.050 \ L = 10 \ mmol$ $n(NaOH) = 0.1 \ M 	imes 0.075 \ L = 7.5 \ mmol$ After reaction: $n(NH_4Cl) = 10 - 7.5 = 2.5 \ mmol$ $n(NH_4OH) = 7.5 \ mmol$ This forms a basic buffer solution containing a weak base $(NH_4OH)$ and its salt $(NH_4Cl)$. Using the Henderson-Hasselbalch equation for a basic buffer: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$ $pOH = 4.74 + \log \frac{2.5}{7.5} = 4.74 + \log \frac{1}{3}$ $pOH = 4.74 - 0.477 = 4.263$ Now,calculate $pH$: $pH = 14 - pOH = 14 - 4.263 = 9.737 \approx 9.74$

Calculate $pH$ of the solution obtained by mixing $50 \ mL$ of $0.2 \ M$ $NH_4Cl$ solution and $75 \ mL$ of $0.1 \ M$ $NaOH$ solution. Given $pK_b$ for aqueous $NH_3$ is $4.74$.

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