Calculate the titration of 25ml of 0.2M maleic acid with 0.2M sodium hydroxide. Calculate the ph of the titration curve using points of the curve a) 0ml b) 12ml c) 25ml d) 37.5ml e) 50ml?

1 Answer
May 6, 2018

WARNING! Very long answer! The pH values are a) 0.78; b) 1.79; c) 3.95; d) 6.07; e) 9.45.

Explanation:

You are titrating a dibasic acid. The equilibria involved are

#"H"_2"A" + "H"_2"O" ⇌ "HA"^"-" + "H"_3"O"^"+"; K_text(a₁) = 1.48 × 10^"-2"; "p"K_text(a₁) = 1.83#
#"HA"^"-" + "H"_2"O" ⇌ "A"^"2-" + "H"_3"O"^"+"; color(white)(ll)K_text(a₂) = 8.51 × 10^"-7"; "p"K_text(a₂) = 6.07#

a) At 0 mL

#color(white)(mmmmmm)"H"_2"A" + "H"_2"O" ⇌ "HA"^"-" + "H"_3"O"^"+"#
#"I/mol·L"^"-1":color(white)(m)0.2color(white)(mmmmmmll)0color(white)(mmml)0#
#"C/mol·L"^"-1":color(white)(m)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mm)"+"x#
#"E/mol·L"^"-1":color(white)(l)"0.2 -" xcolor(white)(mmmmmm)xcolor(white)(mmm)x#

#K_"a" = (["H"_3"O"^+]["A"^"-"])/(["HA"]) = x^2/(0.2-x) = 1.48 × 10^"-2"#

Check for negligibility

#0.2/(1.48 × 10^"-2") = 13 < 400#. ∴ We must solve a quadratic.

#x^2 = (0.2-x) × 0.0148 = 0.0296 - 0.0148x#

#x^2 + 0.0148x - 0.0296 = 0#

#x = 0.165#

#["H"_3"O"^"+"] = xcolor(white)(l) "mol/L" = "0.165 mol/L"#

#"pH" = -log["H"_3"O"^"+"] = -log0.165 = 0.78#

b) At 12 mL

#"Initial moles of acid" = 25 color(red)(cancel(color(black)("mL H"_2"A"))) × ("0.2 mmol H"_2"A")/(1 color(red)(cancel(color(black)("mL H"_2"A")))) = "5.0 mmol H"_2"A"#

#"Moles of NaOH added" = 12color(red)(cancel(color(black)("mL NaOH"))) × ("0.2 mmol NaOH")/(1 color(red)(cancel(color(black)("mL NaOH")))) = "2.4 mmol NaOH"#

#color(white)(mmmmmm)"H"_2"A" + "OH"^"-" ⇌ "HA"^"-" + "H"_2"O"#
#"I/mmol":color(white)(mll)50color(white)(mmll)24color(white)(mmml)0#
#"C/mmol":color(white)(m)"-24"color(white)(mml)"-24"color(white)(mm)"+24"#
#"E/mmol":color(white)(ml)26color(white)(mmm)0color(white)(mmml)24#

#"pH = p"K_text(a₁) + log((["HA"^"-"])/(["H"_2"A"])) = 1.83 + log(24/26) = 1.83 - 0.0348 = 1.79#

c) At 25 mL

#"Moles of NaOH added" = 25color(red)(cancel(color(black)("mL NaOH"))) × ("0.2 mmol NaOH")/(1 color(red)(cancel(color(black)("mL NaOH")))) = "50 mmol NaOH"#

You have neutralized all the #"H"_2"A"#. You are at the first equivalence point.

#"pH" = 1/2("p"K_text(a₁) + "p"K_text(a₂)) = 1/2(1.83 + 6.07) = 7.90/2 =3.95#

d) At 37.5 mL

#"Moles of NaOH added" = 37.5color(red)(cancel(color(black)("mL NaOH"))) × ("0.2 mmol NaOH")/(1 color(red)(cancel(color(black)("mL NaOH")))) = "7.5 mmol NaOH"#

You have neutralized all the #"H"_2"A"# and half the #"HA"^"-"#.

Thus,

#"pH = p"K_text(a₂) = 6.07#

e) At 50 mL

#"Moles of NaOH added" = 50color(red)(cancel(color(black)("mL NaOH"))) × ("0.2 mmol NaOH")/(1 color(red)(cancel(color(black)("mL NaOH")))) = "10 mmol NaOH"#

You have neutralized all the #"H"_2"A"# and all the #"HA"^"-"#.

You have a solution of 5.0 mmol of #"A"^"2-"# in 75 mL.

#["A"^"-"] = "5.0 mmol"/"75 mL" = "0.067 mol/L"#

#color(white)(mmmmmm)"A"^"2-" +color(white)(m) "H"_2"O"color(white)(m) ⇌ color(white)(m)"HA"^"-" + "OH"^"-"#
#"I/mol":color(white)(mll)0.067color(white)(mmmmmmmmm)0color(white)(mmm)0#
#"C/mol":color(white)(mm)"-"xcolor(white)(mmmmmmmmml)"+"xcolor(white)(mm)"+"x#
#"E/mol":color(white)(m)"0.067-"xcolor(white)(mmmmmmmml)xcolor(white)(mmll)x#

#K_"b" = K_"w"/K_"a₂" = (1.00 × 10^"-14")/(8.51 × 10^"-7") = 1.18 × 10^"-8"#

#K_"b" = (["HA"^"-"]["OH"^"-"])/(["A"^"2-"]) = x^2/(0.067-x) = 1.18 × 10^"-8"#

Check for negligibility

#0.067/(1.18 ×10^"-8") = 5.7 × 10^"-6" > 400#. ∴ #x ≪ 0.067#

#x^2 = 0.067 × 1.18 × 10^"-8" = 7.8 × 10^"-10"#

#x = sqrt(7.8 × 10^"-10") = 2.8 × 10^"-5"#

#["OH"^"-"] = 2.8 × 10^"-5" color(white)(l)"mol/L"#

#"pOH" = -log(2.8 × 10^"-5") = 4.55#

#"pH = 14.00 - pOH = 14.00 - 4.55 = 9.45"#

Your calculated values should match the red dots in the titration curve below.

Titration