Start by calculating #"K"_a#, the acid dissociation constant, for pivalic acid from the given conditions. A #"RICE"# (reaction, initial, change, and equilibrium) table might help with the calculation. Let aqueous pivalate anions be represented by #"A"^(-)(aq)#:
Initially- i.e., before #"HA"# ionizes-
#["HA"]_"initial"=0.100color(white)(l)"mol"*"dm"^(-3)#
The equilibrium concentration of #"H"^+# can be calculated from #"pH"#:
#["H"^+]_"equilibrium"=10^(-"pH")=1.0*10^(-3)color(white)(l)"mol"*"dm"^(-3)#
#color(grey)"R"" " "HA" (aq)\rightleftharpoons "H"^(+)(aq) + "A"^(-)(aq)#
#color(grey)("I")color(white)(-lll)0.100#
#color(grey)("C")color(white)(lll)color(grey)(ul(-0.001))color(white)(l-)color(grey)(ul(+0.001))color(white)(lll)color(grey)(ul(+0.001))#
#color(grey)("E")color(white)(-ll)color(grey)(ul(0.099)) color(white)(--)0.001color(white)(--)color(grey)(ul(0.001))#
Fill out missing blanks (as shown in grey) with reference to the stoichiometric relationship between terms in the #color(grey)("C")# row of changes in concentration; Thus
#"K"_a=(["H"^+]*["A"^-])/(["HA"])=1.0*10^(-5)#
As an ionic compound, sodium pivalate would dissociate completely when dissolved in water:
#["A"^-]_"initial"=["Na"^+]=["NaA"]=0.100color(white)(l)"mol"*"dm"^(-3)#
Let the hydroxide concentration at equilibrium,
#["OH"]_"equilibrium"#,
be #x color(white)(l)"mol"*"dm"^(-3)#
The pivalate ion would undergo hydrolysis, as shown in the reversible reaction
#color(grey)"R"" " "A"^(-) (aq)+"H"_2"O"(l)\rightleftharpoons "HA"(aq) + "OH"^(-)(aq)#
#color(grey)("I")color(white)(-lll)0.100#
#color(grey)("C")color(white)(-lll)color(grey)(-x)color(white)(----l--)color(grey)(+x)color(white)(--lll)color(grey)(+x)#
#color(grey)("E")color(white)(-ll)color(grey)(0.100-x) color(white)(---lll-)color(grey)(x)color(white)(-l-l-)color(grey)(x)#
Pivalate ions accept protons from water, and therefore behave as Brownsted-Lowry Base in the forward reaction; Therefore
#"K"_b=(["HA"]*["OH"^-])/(["A"^-])=x^2/(0.100-x) ~~ x^2/0.100#
#"K"_b="K"_w/"K"_a# for conjugate acid-base pairs when the temperature is held constant; assuming #"T"=298color(white)(l)"K"# and #"K"_w=1.0*10^(-14)#,
#"K"_b=1.0*10^(-9)#
Thus #x^2/(0.100)~~1.0*10^(-9)# hence
#["OH"^-]=x~~1.0*10^(-4) color(white)(l)"mol"*"dm"^(-3)#
Again, assuming #"K"_w=1.0*10^(-14)#
#"pH"="pKw"+log["OH"^-]=10.00#
