The titration takes place in two stages:
In Stage 1, you are neutralizing the #"CO"_3^"2-"#:
#"CO"_3^"2-" + "H"_3"O"^"+" → "HCO"_3^"-" + "H"_2"O"#
In Stage 2, you are neutralizing the #"HCO"_3^"-"#:
#"HCO"_3^"-" + "H"_3"O"^"+" → "H"_2"CO"_3 + "H"_2"O"#
(a) pH at 30.5 mL #"HCl"#
At this point, you have used 25 mL of #"HCl"# to reach the first equivalence point and have added 5.5 mL of #"HCl"# towards the second equivalence point.
#color(white)(mmmmmll)"HCO"_3^"-" + "H"_3"O"^"+" → "H"_2"CO"_3 + "H"_2"O"#
#"I/mmol":color(white)(mll)2.50color(white)(mm)0.550color(white)(mmmll)0#
#"C/mmol":color(white)(ll)"-0.550"color(white)(ml)"-0.550"color(white)(mm)"+0.550"#
#"E/mmol":color(white)(ml)1.95color(white)(mmm)0color(white)(mmmm)0.550#
#"Initial moles HCO"_3^"-" = 0.025 color(red)(cancel(color(black)("L HCl"))) × (0.100 color(red)(cancel(color(black)("mol HCl"))))/(1 color(red)(cancel(color(black)("L HCl")))) × "1 mol HCO"_3^"-"/(1 color(red)(cancel(color(black)("mol HCl")))) = "0.002 50 mol HCO"_3^"-" = "2.50 mmol HCO"_3^"-"#
#"Added moles HCl" = 0.0055 color(red)(cancel(color(black)("L HCl"))) × "0.100 mol HCl"/(1 color(red)(cancel(color(black)("L HCl")))) = "0.000 550 mol HCl" = "0.550 mmol HCl"#
We have a buffer consisting of 1.95 mmol #"HCO"_3^"-"# and 0.550 mol #"H"_2"CO"_3#.
According to the Henderson-Hasselbalch equation,
#color(blue)(bar(ul(|color(white)(a/a)"pH" = "p"K_text(a) + log(["HCO"_3^"-"]/["H"_2"CO"_3])color(white)(a/a)|)))" "#
For the first ionization of #"H"_2"CO"_3#, #K_text(a) = 4.27 × 10^"-7";"p"K_text(a) = 6.37#.
∴ #"pH" = 6.37 + log(0.550/1.95) = "6.37 + log0.282" = "6.37 - 0.550" = 5.82#
pH at second equivalence point
At this stage we have added 50 mL of #"HCl"# and have converted all of the #"CO"_3^"2-"# to #"H"_2"CO"_3#.
#"Amt. of H"_2"CO"_3 = "2.50 mmol"#
#"Total volume" = "(25 + 50) mL" = "75 mL"#
#["H"_2"CO"_3] = "2.50 mmol"/"75 mL" = "0.0333 mol/L"#
#color(white)(mmmmmml)"H"_2"CO"_3 + "H"_2"O" ⇌ "HCO"_3^"-" + "H"_3"O"^"+"#
#"I/mol·L"^"-1":color(white)(ml)0.0333color(white)(mmmmmmm)0color(white)(mmmll)0#
#"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmmml)"+"xcolor(white)(mmm)"+"x"#
#"E/mol·L"^"-1":color(white)(ll)"0.0333-"xcolor(white)(mmmmmml)xcolor(white)(mmmll)x#
#K_text(a) = (["H"_3"O"^"+"]["HCO"_3^"-"])/(["H"_2"CO"_3]) = x^2/(0.0033-x) = 4.27 × 10^"-7"#
#0.0333/(4.27 × 10^"-7") = 7.81 × 10^4. ∴ x ≪0.0500#
Then,
#x^2 = 0.0333 × 4.27 × 10^"-7" = 1.42 × 10^"-8"#
#x = 1.19 × 10^"-4"#
#["H"_3"O"^"+"] = 1.19 × 10^"-4"color(white)(l) "mol/L"#
#"pH" = -log["H"_3"O"^"+"] = -log(1.19 × 10^"-4") = 3.92#
