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Question #c1b5b

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Answer 1 · 2016-11-21T13:54:56

The pH of a buffer solution depends on the ratio of concentrations of an acid and its conjugate base, using the Henderson-Hasselbalch equation.
$p H = p K_{a} - {log}_{10} (\frac{[H A]}{[A^{-}]})$ $pH=pK_a-log_10(([HA])/([A^-]))$

Explanation:

In this example, the acid is $N H_{4}^{+}$ $NH_4^+$ , which has a $p K_{a} = 9.24$ $pK_a=9.24$

Total volume of solution = 30mL + 15 mL = 45 mL = 0.45 L

Acid concentration: $\frac{0.1 M \times 0.015 L}{0.045 L} = 0.033 M$ $(0.1M times 0.015L)/(0.045L) = 0.033M$

Base concentration: $\frac{0.2 M \times 0.030 L}{0.045 L} = 0.133 M$ $(0.2M times 0.030L)/(0.045L) = 0.133M$

$p H = p K_{a} - {log}_{10} (\frac{[H A]}{[A^{-}]}) = 9.24 - {log}_{10} (\frac{0.033}{0.133}) = 9.84$ $pH=pK_a-log_10(([HA])/([A^-]))=9.24-log_10(0.033/0.133)=9.84$

Sanity Check:
If the acid and base concentrations were equal, the $p H$ $pH$ would be equal to the $p K_{a}$ $pK_a$ , but here the base concentration is 4X the acid concentration, so the solution is a little bit more basic and the $p H$ $pH$ is a little bit higher than $p K_{a}$ $pK_a$ .

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