Step 1. Calculate the concentration of the weak acid
We can use the dilution formula:
#color(blue)(bar(ul(|color(white)(a/a)c_1V_1 = c_2V_2color(white)(a/a)|)))" "#
The formula gives
#c_2 = c_1 × V_1/V_2#
#c_1 = "0.125 mol/L"; V_1 = color(white)(ll)"50 mL"#
#c_2 = ?; color(white)(mmmmml)V_2 = "115 mL"#
∴ #c_2 = "0.125 mol/L" × (50 color(red)(cancel(color(black)("mL"))))/(115 color(red)(cancel(color(black)("mol/L")))) = "0.0543 mol/L"#
Step 2. Calculate the pH
The chemical equation is
#"HA" + "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-" ; "p"K_text(a) = 4.15"#
We can use an ICE table here.
#color(white)(mmmmmmmmll)"HA" + "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-"#
#"I/mol·L"^"-1":color(white)(mmll)0.0543color(white)(mmmmmm)0color(white)(mmm)0#
#"C/mol·L"^"-1":color(white)(mmmll)"-"xcolor(white)(mmmmmm)"+"xcolor(white)(mml)"+"x#
#"E/mol·L"^"-1":color(white)(mm
)"0.0543 -"xcolor(white)(mmmmm)xcolor(white)(mmm)x#
#K_text(a) = 10^("-"4.15) = 7.08 × 10^"-5"#
#K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = (x × x)/("0.0543 -"color(white)(l)x) = x^2/("0.0543 -"color(white)(l)x) = 7.08 × 10^"-5""#
Check for negligibility:
#0.0543/(7.080 × 10^"-5") = 768 > 400#.
∴ #x ≪ 0.0543#
Then
#x^2/0.0543 = 7.08 × 10^"-5"#
#x^2= 0.0543× 7.08 × 10^"-5" = 3.85 ×10^"-6"#
#x = 1.96 × 10^"-3"#
Step 4. Calculate the pH
#["H"_3"O"^"+"] = x color(white)(l)"mol/L" = 1.96 × 10^"-3"color(white)(l)"mol/L"#
#"pH" = "-log"["H"_3"O"^"+"] = "-log"(1.96 × 10^"-3") = 2.71#
