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?

#["NaOH"] = "2.00 M"#

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"#