2. What is the pH of a buffer made up of 0.5M NH, and 0.456M NH4CI? pKb NH₂ = 4.74
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2. What is the pH of a buffer made up of 0.5M NH, and 0.456M NH4CI? pKb NH₂ = 4.74
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[tex]NH_4 \leftrightharpoons NH_3^+ +H^+ \\\\\begin{matrix} & NH_4 & \leftrightharpoons & NH_3^+ & + & H^+ \\\text{initial} & 0.456 & & 0.5 & & 0 \\\text{change} & -x && x && x\\ \text{equilibrium} & 0.456-x &&0.5-x&&x\end{matrix}[/tex]
[tex]pK_b=-log{K_b} \\K_b=10^{-4.74} \\K_w=K_a\cdot K_b\\K_a=\frac{K_w}{K_b} = \frac{1\times 10^{-14}}{10^{-4.74}}=5.4954\times 10^{-10}[/tex]
[tex]K_a = \frac{[H^+][NH_3^+]}{[NH_4]}=\frac{(x)(0.5-x)}{0.456-x} = 5.4954\times 10^{-10}\\x=[H^+]=5.0118\times 10^{-10} M\\pH=-log[H^+]=9.3[/tex]
Using the Henderson-Hasselbalch equation:
[tex]pH=pK_a+log\frac{[A^-]}{[HA]}[/tex]
[tex]pH=(14-4.74)+log\frac{0.5}{0.456}=9.3[/tex]