Warning! Long Answer.
I assume that you have titrated to the end point.
The equation for the reaction is
#"HA" + "KOH" → "KA" +"H"_2"O"#
Since #"HA"# and #"KOH"# react in a 1:1 molar ratio, we can write
#c_aV_a =c_bV_b#
∴ #c_a = c_b × V_b/V_a = "0.40 mol/L" × (26 color(red)(cancel(color(black)("drops"))))/(12 color(red)(cancel(color(black)("drops")))) = "0.867 mol/L"#
At neutralization, you have only #"KA"# present.
If there had been no change in volume, its concentration would have been 0.867 mol/L.
However, the volume increased during the titration to #"(12 drops + 26 drops) = 38 drops"#.
To calculate the new concentration, we can use the dilution formula
#color(blue)(|bar(ul(color(white)(a/a) c_1V_1 = c_2V_2color(white)(a/a)|)))" "#
#c_1 = "0.867 mol/L"; V_1 = "12 drops"#
#c_2 = ?; color(white)(mmmmml)V_2 = "38 drops"#
#c_2 = c_1 × V_1/V_2 = "0.867 mol/L" × (12 color(red)(cancel(color(black)("drops"))))/(28color(red)(cancel(color(black)( "drops")))) = "0.372 mol/L"#
∴ At neutralization, we have a 0.372 mol/L solution of #"KA"#.
Calculate the concentration of #["H"^+]# in a 0.372 mol/L solution of KA
#color(white)(mmmmmmmmll)"A"^"-" + "H"_2"O" ⇌ "HA" + "OH"^"-"#
#"I/mol·L"^"-1": color(white)(mml)0.372color(white)(mmmmmll) 0color(white)(mmll) 0#
#"C/mol·L"^"-1":color(white)(mmm) "-"xcolor(white)(mmmmml) +x color(white)(m)+x#
#"E/mol·L"^"-1": color(white)(m)0.372 -xcolor(white)(mmmmll) x color(white)(mmll)x#
#K_b = K_w/K_a = (1.00 × 10^"-14")/(6.46 × 10^"-5") = 1.55 × 10^"-10"#
#K_b = (["HA"]["OH"^"-"])/(["A"^"-"])#
#1.55 × 10^"-10" = (x × x)/(0.372 -x) - x^2/(0.372-x)#
Check if #x# is negligible
#0.372/(1.55 × 10^"-10") = 2.11 × 10^9 > 400; ∴ x « 0.372#
#1.55 × 10^"-10" = x^2/0.372#
#x^2 = 0.372 × 1.55 × 10^"-10" = 5.77 × 10^"-11"#
#x = ["OH"^"-"] = 7.59 ×10^"-6"#
#["H"^+] = K_w/["OH"^"-"] = (1.00 × 10^"-14")/(7.59 ×10^"-6") = 1.3 × 10^"-9"color(white)(l) "mol/L"#
