Question

Part 4: Calculating pH values as a Titration Progresses - this is a tough question! Finally, consider a 20.00 mL aqueous solution of 0.20 M H2A, where "H2A" is some arbitrary diprotic acid (Ka1 = 4.35 x 10-7, Kaz = 4.85 x 10-11 Estimate the pH values of the solution after the addition of 0.00, 10.00, 20.00, 30.00, 40.00, and 60.00 mL of 0.20 M NaOH.

Answer 1

Data V = 20 mL of H.A, 0.2 M Kay= 4.35x10-7 K = 4.85x10-11 Determine the pH: a) Before the addition of any NaOH HA + H2O + 0.2 -X 0.2 - X HA++ H2O* 0 | C E 0 +x x +X x Kat [HA [H,0+] [H_A] 4.35x 10-7 = = 0.2 - x Aprox. 0.2 - x = 0.2 4.354107= 4.35x10"(0.2) = x 2.95x 104 = X ----> [H2O'] pH = -log [H2011 pH = -log 2.95x10-4 pH = 3.5

b) after adding 10 mL of NaOH 0.2 M moles of NaOH added = 0.01 Ex 0.2 mol/L = 0.002 moles Initial moles of H2A = 0.02 Ex 0.2 mol/E = 0.004 moles Vtotal = 30 mL = 0.03 L HA + HO HA+ + H2O 0.004 0.002 -0.002 -0.002 0.002 0.002 [H,A] = 0.002 mol/ 0.03 L = 0.07 M [HA] = 0.002 mol/ 0.03 L = 0.07 M. Ka = 4.35x 10-7 pkg, = -log 4.35x107 pka = 6.36 pH = pky + log [HA] [HA] pH = pka pH = 6.36

c) After adding 20 mL of NaOH 0.2 M moles of NaOH added = 0.02 E x 0.2 mol/E = 0.004 moles Initial moles of H, A = 0.02 EX 0.2 mol/L = 0.004 moles Vtotal = 40 mL = 0.04 L HA + + H2O H.A + HO 0.004 0.004 -0.004 -0.004 Equivalence 0.004 point [HA] = 0.004 moles/ 0.04 L = 0.1 M HA++ H20 + | 0.004 C -X E 0.004-x A²+ H20* 0 0 +x +x x x [HA][H,0*] [H_A] K2= 4.85x10-11 =- x2 0.004 - X Value of x is so small that it can be neglected, 0.004 - X = 0.004 72 4.85x10-11 =- 0.004 4.85x10-11 (0.004) = x 2.64x108 = x 2.64x 10-8 = x ----> [H20*] pH = -log [H20*] pH = -log 2.64x10-8 pH = 7.58

d) After adding 30 mL of NaOH 0.2 M. moles of NaOH added = 0.03 Ex 0.2 mol/E = 0.006 moles Initial moles of H2A = 0.02 EX 0.2 mol/E = 0.004 moles Viotal = 50 mL = 0.05 L - A2- + H,0+ HA+ OH 0.1 0.006 -0.006 -0.006 0.094 0 0.006 [HA] = 0.094 mol/ 0.05 L = 1.88 M [A²] = 0.006 mol/ 0.05 L = 0.12 M. pKq2 = -logka2 pKq2 = -log4.85x10-11 pKq2 = 10.31 [A?] pH = pkan + log [HA] 0.12 pH = pka + log – 1.88 pH = 10.31 - 1.19 pH = 9.12

e) After adding 40 mL of NaOH 0.2 M moles of NaOH added = 0.04 E 0.2 mol/E = 0.008 moles Initial moles of H A = 0.02 Ex 0.2 mol/t = 0.004 moles Vtotal = 60 mL = 0.06 L [HA] = 0.004 moles/ 0.04 L = 0.1 M HA+ oH A2 + H2O 0.1 0.008 -0.008 -0.008 0.092 0 0.008 [HA] = 0.092 mol/ 0.06 L = 1.53 M [A] = 0.008 mol/ 0.06 L = 0.13 M (A? pH = pk, + log 2 * log - [HA] 0.13 pH = pka2 + log 1.53 pH = 10.31 - 1.07 pH = 9.24

f) After adding 60 mL of NaOH 0.2 M moles of NaOH added = 0.06 EX 0.2 mol/E = 0.012 moles Vtotal = 80 mL = 0.08 L - A2- + H,O HA+ OH 0.1 0.012 -0.012 -0.012 0.088 0.012 [HA] = 0.088 mol/ 0.08 L = 1.1 M [A?] = 0.012 mol/ 0.08 L = 0.15 M [A21 pH = pk, + log [HA] 0.15 pH = pk 2 + log 71 pH = 10.31 - 0.85 pH = 9.46