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#?

Question

Mateo Aguilar · Accepted Answer

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).

For the time being, use the Henderson-Hasselbalch equation to determine the ratio of these substances:

#"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.

At the same time, you dilute both to the same degree:

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

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

This results in:

#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#.

Peyton Adams · Answer

undefined Adding distilled water to the buffer solution dilutes the concentration of NH3 and NH4Cl without significantly affecting the concentrations of NH4+ and OH- ions. Therefore, the pH of the solution will increase slightly due to the decrease in the concentration of the buffer components. The change in pH can be calculated using the Henderson-Hasselbalch equation. However, the exact change in pH depends on the initial concentrations of NH3 and NH4Cl in the buffer solution.