A $100.00-mL$ $"100.00-mL"$ solution containing $0.500 M$ $"0.500 M"$ ${NH}_{3}$ $"NH"_3$ and $0.500 M$ $"0.500 M"$ ${NH}_{4}^{+}$ $"NH"_4^+$ ( ${pK}_{a} = 9.26$ $"pK"_a = 9.26$ ) had $NaOH$ $"NaOH"$ added to it, and increased to a $pH$ $"pH"$ of $10.26$ $10.26$ . If $20.45 mL$ $"20.45 mL"$ of $NaOH$ $"NaOH"$ was added, what was its molarity at the time of addition?

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A $100.00-mL$ $"100.00-mL"$ solution containing $0.500 M$ $"0.500 M"$ ${NH}_{3}$ $"NH"_3$ and $0.500 M$ $"0.500 M"$ ${NH}_{4}^{+}$ $"NH"_4^+$ ( ${pK}_{a} = 9.26$ $"pK"_a = 9.26$ ) had $NaOH$ $"NaOH"$ added to it, and increased to a $pH$ $"pH"$ of $10.26$ $10.26$ . If $20.45 mL$ $"20.45 mL"$ of $NaOH$ $"NaOH"$ was added, what was its molarity at the time of addition?

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Answer 1 · 2018-02-24T04:13:02

$[NaOH] = 2.00 M$ $["NaOH"] = "2.00 M"$

The first thing we can do is find out what the ratio of weak acid to weak base turns out to be. The Henderson-Hasselbalch equation applies because ${NH}_{4}^{+}$ $"NH"_4^+$ is the conjugate acid of ${NH}_{3}$ $"NH"_3$ , thereby forming a buffer.

$pH = {pK}_{a} + log \frac{[{NH}_{3}]}{[{NH}_{4}^{+}]}$ $"pH" = "pK"_a + log\frac(["NH"_3])(["NH"_4^+])$

$10.26 = 9.26 + log \frac{[{NH}_{3}]}{[{NH}_{4}^{+}]}$ $10.26 = 9.26 + log\frac(["NH"_3])(["NH"_4^+])$

$1.00 = log \frac{[{NH}_{3}]}{[{NH}_{4}^{+}]}$ $1.00 = log\frac(["NH"_3])(["NH"_4^+])$

Therefore, $[{NH}_{3}] / [{NH}_{4}^{+}] = 10$ $["NH"_3]"/"["NH"_4^+] = 10$ . Now, this is after adding $20.45 mL NaOH$ $"20.45 mL NaOH"$ , which adds a certain number of $mols$ $"mols"$ . Call that $x$ $x$ .

The strong base $NaOH$ $"NaOH"$ reacts fully with the weak acid, neutralizing it and forming an equal number of $mols$ $"mols"$ of weak base ${NH}_{3}$ $"NH"_3$ . Thus,

$10 = \frac{[{NH}_{3}]}{[{NH}_{4}^{+}]} = \frac{0.500 M \cdot 100.00 \times 10^{- 3} L + x}{0.500 M \cdot 100.00 \times 10^{- 3} L - x}$ $10 = \frac(["NH"_3])(["NH"_4^+]) = ("0.500 M" cdot 100.00 xx 10^(-3) "L" + x)/("0.500 M" cdot 100.00 xx 10^(-3) "L" - x)$

$= \frac{0.0500 + x}{0.0500 - x}$ $= (0.0500 + x)/(0.0500 - x)$

This sets up an equation to solve for the $mols$ $"mols"$ of $NaOH$ $"NaOH"$ .

$10 (0.0500 - x) = 0.0500 + x$ $10(0.0500 - x) = 0.0500 + x$

$0.500 - 10 x = 0.0500 + x$ $0.500 - 10x = 0.0500 + x$

$11 x = 0.500 - 0.0500$ $11x = 0.500 - 0.0500$

$=> x = (0.450)/(11) = ul"0.0409 mols NaOH"$

As a result, by knowing that we added $"20.45 mL"$ of $"NaOH"(aq)$ , its concentration was:

$color(blue)(["NaOH"]) = "0.0409 mols"/(20.45 cancel"mL") xx (1000 cancel"mL")/("1 L")$

$=$ $ulcolor(blue)"2.00 M"$

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I wrote this question as an exam review for my students. :)

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