Question

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

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

Answer 1

`["NaOH"] = "2.00 M"`

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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^+` is the conjugate acid of `"NH"_3`, thereby forming a buffer.

`"pH" = "pK"_a + log\frac(["NH"_3])(["NH"_4^+])`

`10.26 = 9.26 + log\frac(["NH"_3])(["NH"_4^+])`

`1.00 = log\frac(["NH"_3])(["NH"_4^+])`

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

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

`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)`

`= (0.0500 + x)/(0.0500 - x)`

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

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

`0.500 - 10x = 0.0500 + x`

`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"`