# 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?

Jul 19, 2017

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

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

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

The $\text{pKa}$ always belongs to the acid, and the $\text{pKa}$ of ${\text{NH}}_{4}^{+}$ is about $9.26$. Since $\text{pH" = "pKa}$, we already know that

$\left[{\text{NH"_3] = ["NH}}_{4}^{+}\right]$,

since

$9.26 = 9.26 + \log \setminus \frac{\left[{\text{NH"_3])(["NH"_4^(+)]) => \frac(["NH"_3])(["NH}}_{4}^{+}\right]}{=} 1$

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

Since we know that $\left[{\text{NH"_3] = ["NH}}_{4}^{+}\right]$, and these substances are $1 : 1$ in the equilibrium

${\text{NH"_3(aq) + "H"_2"O"(l) rightleftharpoons "NH"_4^(+)(aq) + "OH}}^{-} \left(a q\right)$,

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

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

$\left[{\text{NH"_3] -> 1/2["NH}}_{3}\right]$

$\left[{\text{NH"_4^(+)] -> 1/2["NH}}_{4}^{+}\right]$

This gives:

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

$= 9.26 + \log \left(1\right) = \textcolor{b l u e}{\text{pH} = 9.26}$

as before. So the $\text{pH}$ doesn't change from $9.26$.