Question

A buffer solution of `NH_3` and `NH_4Cl` has a `pH = 9.26`. If in 100mL of buffer solution, 100 mL of distilled water is added what is the change of `pH`?

Answer 1

Zero. All you've done is double the volume. The ratio of concentrations stays the same. If a buffer `"pH"` was dependent on its volume, then `"pH"` would not be intensive (but it is intensive).

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For now, calculate the ratio of these substances using the Henderson-Hasselbalch equation:

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

The `"pKa"` always belongs to the acid, and the `"pKa"` of `"NH"_4^(+)` is about `9.26`. Since `"pH" = "pKa"`, we already know that

`["NH"_3] = ["NH"_4^(+)]`,

since

`9.26 = 9.26 + log\frac(["NH"_3])(["NH"_4^(+)]) => \frac(["NH"_3])(["NH"_4^(+)]) = 1`

If the buffer solution started off at `"100 mL"`, a two-fold dilution to approximately `"200 mL"` total volume (when assuming solution volumes are additive!) should do nothing to the `"pH"`.

Since we know that `["NH"_3] = ["NH"_4^(+)]`, and these substances are `1:1` in the equilibrium

`"NH"_3(aq) + "H"_2"O"(l) rightleftharpoons "NH"_4^(+)(aq) + "OH"^(-)(aq)`,

there are `"50 mL"` of each in a `1:1` molar ratio. So, I suppose this is a two-fold dilution from `"50 mL"` to `"100 mL"` of each, but it doesn't matter exactly.

You dilute both to the same extent, at the same time:

`["NH"_3] -> 1/2["NH"_3]`

`["NH"_4^(+)] -> 1/2["NH"_4^(+)]`

This gives:

`color(blue)("pH"') = 9.26 + log\frac(cancel(1/2)["NH"_3])(cancel(1/2)["NH"_4^(+)])`

`= 9.26 + log(1) = color(blue)("pH" = 9.26)`

as before. So the `"pH"` doesn't change from `9.26`.