#K_b -= K_w / K_a = 1.00 xx 10^(-8)#
The strong base #"NaOH"# is in excess and neutralizes all #"HA"# initially present in the solution. As a result, the hydrolysis of #"A"^(-)#, the product of the neutralization reaction. would be the only reversible process capable of generating an equilibrium.
#"A"^(-)(aq) + "H"_2"O"(aq)\
color(purple)(rightleftharpoons) \
"HA"(aq) + ul("OH"^(-) (aq))#
R #"A"^(-)(aq) + "H"_2"O"(aq)\
color(purple)(rightleftharpoons) \
"HA"(aq) + ul("OH"^(-) (aq))#
I #0.005 color(white)(------------) 0.005#
C #-x color(white)(-------ll)+x color(white)(----) +x#
E #0.005 -x color(white)(-----l-)x color(white)(---ll) 0.005 +x#
(in #mol#)
#c = n /color(purple)(V)# and therefore
#K_b("A"^(-)) = (overbrace(["HA"])^("conjugate acid") * ["OH"^(-)])/(overbrace(["A"^(-)])^("weak base"))\
= ((x)/(0.150)*(0.005+x))/(0.005 -x )#
#(x*(0.005+x))/(0.005 -x ) = 1.00 xx 10 ^(-8) * 0.150#
#x^2 + (0.005 + 1.00 xx 10^(-8))* x -0.005 * 1.00 xx 10 ^(-8) = 0#
#x = 1.50 xx 10 ^(-9) color(white)(l) mol#
(#x > 0# given that #["HA"] >= 0#)
Therefore
#n("OH"^(-)) = 0.005 + 1.50 * 10^(-9) ~~ 0.005 color(white)(l) mol#
#c("OH"^(-)) = n / V = 0.0333 color(white)(l) mol * dm^(-3)#
#"pH" = "pKw" - "pOH" = 12.523#
Note that #"OH"^(-)#, already present at great quantity, is the only factor through which the equilibrium influences the #"pH"# of the solution. #K_b = 1.00 xx 10^(-8)#; thus there's a small deviation in #"pH"# from what would have been observed with a strong acid of the same volume and concentration.
