pH of solution
To a first approximation, we can assume that only the first ionization is important in determining the pH.
#color(white)(mmmmmmm)"H"_2"A" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HA"^"-"#
#"I/mol/L:"color(white)(mm)0.0650color(white)(mmmmmml)0color(white)(mmml)0#
#"C/mol/L:"color(white)(mmm)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mmll)"+"x#
#"E/mol/L:"color(white)(ml)"0.0650-"xcolor(white)(mmmmmll)xcolor(white)(mmml)x#
#K_text(a1) = (["H"_3"O"^"+"]["HA"^"-"])/(["H"_2"A"]) = x^2/("0.0650-"x) = 2.4 × 10^"-6"#
#0.0650/(2.4 × 10^"-6") = 2.7 × 10^4 ≫ 400. ∴ x ≪0.0650#
Then
#x^2/0.0650 = 2.4 × 10^"-6"#
#x^2 = 0.0650 × 2.4 × 10^"-6" = 1.56 × 10^"-7"#
#x = 3.95 × 10^"-4"#
#["H"_3"O"^"+"] = ["HA"^"-"] = 3.95 × 10^"-4" color(white)(l)"mol/L"#
#"pH" = -log["H"_3"O"^"+"] = -log(3.95 × 10^"-4") = 3.40#
Equilibrium concentration of #"H"_2"A"#
#["H"_2"A"] = "(0.0650-"x") mol/L" = (0.0650 - 3.95×10^"-4"") mol/L" = "0.0646 mol/L"#
Equilibrium concentration of #"A"^"2-"#
For this, we can use the second ionization step.
#"HA"^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"2-"#
#K_text(a2) = (["H"_3"O"^"+"]["A"^"2-"])/(["HA"^"-"])#
#["A"^"2-"] = (K_text(a2)["HA"^"-"])/(["H"_3"O"^"+"]) = (8.7 × 10^"-9" × color(red)(cancel(color(black)(3.95 × 10^"-4"))))/(color(red)(cancel(color(black)(3.95 × 10^"-4")))) = 8.7 × 10^"-9"color(white)(l) "mol/L"#
