Consider a weak acid #"HA"#, where #"A"^(-)# is some arbitrary anion and #"HA"(aq) + "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "A"^(-)(aq)#. Suppose you add some #"NaA"(s)# to #"H"_2"O"(l)#. What is the pH of the solution?

To understand this effect, one should calculate the %Ionization of a weak monoprotic acid in pure water and compare it to the %Ionization when the weak acid is in the presence of a quantity of #A^-.#

Assume the hypothetical problems...
(a) determine the %Ionization and pH of a 0.10M monoprotic acid (HA) in pure water. #K_a=1xx10^-5#, and
(b) determine the %Ionization and pH of a 0.10M monoprotic acid (HA) in a solution containing 0.01M NaA.

%Ionization of HA in pure water:
#color(white)(mmmmm)HA##color(white)(m)rightleftharpoonscolor(white)(mm)H^+color(white)()+color(white)(m)A^-#
#C_i:color(white)(mm)0.10M##color(white)(mmmm)0.00Mcolor(white)(m)0.00M#

#DeltaC:color(white)(m)-x##color(white)(mmmmmm)+xcolor(white)(m)+x#

#C_(eq):color(white)(mm)(0.10-x)M##color(white)(mm)xcolor(white)(mmm)x#
#color(white)(mmmm)~~0.10M#

#K_a=([H^+][A^-])/([HA])#=#(x^2)/(0.10)=1xx10^-5#

=> #x=[H^+]=sqrt((0.10)(1xx10^-5))=1xx10^-3M#
=> #%"Ionization"#=#([H^+])/([HA])xx100%=(10^-3)/(10^-1)xx100%=1.0%#
=> #pH=-log[H^+]=-log(10^-3)=3#

%Ionization of HA in 0.01M#A^-#:
#color(white)(mmmm)HA##color(white)(mmm)rightleftharpoonscolor(white)(mmm)H^+color(white)()+color(white)(m)A^-#
#C_i:color(white)(m)0.10M##color(white)(mmmmmm)0.00Mcolor(white)(mm)0.010M#

#DeltaC:color(white)(m)-x##color(white)(mmmmmm)+xcolor(white)(mmmm)+x#

#C_(eq):color(white)(mm)(0.10-x)M##color(white)(mm)xcolor(white)(mmm)(0.010+x)M#
#color(white)(mmmmm)~~0.10color(white)(mmmmmmmm)~~0.01M#

#K_a=([H^+][A^-])/([HA])=(x(0.01M))/(0.10M)=1xx10^-5#

=> #x=[H^+]=((0.1M)(1xx10^-5))/(0.01M)=1xx10^-4M#
=> #%"Ionization"#=#([H^+])/([HA])xx100%=(10^-4)/(10^-1)xx100%=0.10%#
=> #pH=-log[H^+]=-log(10^-4)=4#

So, in pure water pH HA => 3 but in presence of common ion (#A^-#) increases to 4 b/c of decreased ionization of HA. However, both systems remain pH < 7 (acidic).

When you dissolve the weak acid #"HA"# in water, you get the equilibrium reaction.

#underbrace("HA(aq)")_color(red)("acid") + "H"_2"O(l)" ⇌ underbrace("A"^("-")"(aq)")_color(red)("base") + "H"_3"O"^("+")"(aq)"#

We could have written the equilibrium as

# underbrace("A"^("-")"(aq)")_color(red)("base") + "H"_3"O"^("+")"(aq)" ⇌ underbrace("HA(aq)")_color(red)("acid") + "H"_2"O(l)"#

The base removes a proton from the hydronium ion and forms water.

The base can remove a proton from water and form hydroxide ions.

# underbrace("A"^("-")"(aq)")_color(red)("base") + "H"_2"O(l)" ⇌ underbrace("HA(aq)")_color(red)("acid") + "OH"^"-""#

Thus, an aqueous solution of the salt of a weak acid is basic and the pH is greater than 7.

EXAMPLE

If the weak acid has #K_text(a) = 1.00 × 10^"-5"#, what is the pH of a 0.100 mol/L solution of #"NaA"#?

Solution

We can use an ICE table to help with the calculation.

#color(white)(mmmmmmmll)"A"^"-" + "H"_2"O" ⇌ "HA" + "OH"^"-"#
#"I/mol·L"^"-1":color(white)(mll)0.100color(white)(mmmmmll)0color(white)(mmll)0#
#"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmm)"+"xcolor(white)(mll)"+"x#
#"E/mol·L"^"-1":color(white)(m)"0.100-"xcolor(white)(mmmmm)xcolor(white)(mmm)x#

#K_text(b) = K_text(w)/K_text(a) = (1.00 × 10^"-14")/(1.00 × 10^"-5") = 1.00 × 10^"-9"#

#K_text(b) = (["HA"]["OH"^"-"])/(["A"^"-"]) = x^2/("0.100-"x) = 1.00 × 10^"-9"#

Apply the 5 % rule

#1.00/(1.00 × 10^"-9") = 1.00 × 10^9 > 400#.

∴ #x ≪1.00#

Then

#x^2/0.100 = 1.00 × 10^"-9"#

#x^2 = 0.100(1.00 × 10^"-9") = 1.00 × 10^"-10"#

#x = 1.00 × 10^"-5"#

#["OH"^"-"] = x color(white)(l)"mol/L" = 1.00 × 10^"-5"color(white)(l) "mol/L"#

#"pOH" = "-log"["OH"^"-"] = "-log"(1.00 × 10^"-5") = 5.00#

#"pH = 14.00 - pOH = 14.00 - 5.00 = 9.00"#