Our first task is to calculate the concentration of the potassium benzoate,
#"Molarity" = "moles"/"litres" = "0.600 mol"/"1.50 L" = "0.400 mol/L"#
The reaction is
#"C"_7"H"_5"O"_2^"-" + "H"_2"O" ⇌ "HC"_7"H"_5"O"_2 + "OH"^"-"#
For benzoic acid, #K_a = 6.25 × 10^"-5"#.
For the benzoate ion, as above, #K_"b" = K_"w"/K_"a" = (1.00 × 10^"-14")/(6.25 × 10^"-5") = 1.60 × 10^"-10"#
Now, we can set up an ICE table to calculate the concentrations.
#color(white)(mmmmmm)"C"_7"H"_5"O"_2^"-" + "H"_2"O" ⇌ "HC"_7"H"_5"O"_2 + "OH"^"-"#
#"I/mol·L"^"-1": color(white)(ml)0.400color(white)(mmmmmmmmm)0color(white)(mmmll)0#
#"C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmmmmmmll)+xcolor(white)(mm)+x#
#"E/mol·L"^"-1": color(white)(ll)"0.400 -" xcolor(white)(mmmmmmmml)xcolor(white)(mmml)x#
#K_"b" = (["HC"_7"H"_5"O"_2]["OH"^"-"])/(["C"_7"H"_5"O"_2^"-"]) = (x × x)/(0.400 -x) = x^2/(0.400 - x) = 1.60 × 10^"-10"#
#0.400/(1.60 × 10^"-10") = 2.50 × 10^9#
∴ #x ≪ 0.400#
The equation reduces to
#x^2/0.400 = 1.60 × 10^"-10"#
#x^2 = 0.400 × 1.60 × 10^"-10" = 6.40 × 10^"-11"#
#x = 8.00 × 10^"-6"#
∴ #["OH"^"-"] = 8.00 × 10^"-6"color(white)(l) "mol/L"#
#"pOH" = -log["OH"^"-"] = -log(8.00 × 10^"-6") = 5.10#
#"pH" = "14.00 - pOH" = "14.00 - 5.10" = 8.90#
