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#?
1 Answer
Zero. All you've done is double the volume. The ratio of concentrations stays the same. If a buffer
For now, calculate the ratio of these substances using the Henderson-Hasselbalch equation:
#"pH" = "pKa" + log\frac(["NH"_3])(["NH"_4^(+)])#
The
#["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
Since we know that
#"NH"_3(aq) + "H"_2"O"(l) rightleftharpoons "NH"_4^(+)(aq) + "OH"^(-)(aq)# ,
there are
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