The following questions are based on the ionization of carbonic acid. H_2CO_3 hA

Question

The following questions are based on the ionization of carbonic acid. H_2CO_3 hArr H^+ + HCO_3^- K_1 = 4.7 times 10^-7 pK_1 = 6.34 HCO_3^- hArr H^+ + CO_3^-2 K_2 = 4.4 times 10^-11 pK_2 = 10.36 Calculate the pH of a solution formed by mixing the following: 50.0 ml of 0.100 M H_2CO_3 with 50.0 ml of 0.050 M NaOH. 50.0 ml of 0.100 M H_2CO_3 with 50.0 ml of 0.150 M NaOH. 50.0 ml of 0.100 M H_2CO_3 with 50.0 ml of 0.100 M NaOH. 50.0 ml of 0.100 M H_2CO_3 with 50.0 ml of 0.200 M NaOH. Electrochemistry A solution contains: 0 100 M Cc^+3; 0 100 times 10^-4 M Ce^+4; 0.100 times 10^-4 M Mn^+2: 0.100M MnO_4^-1; and 1.00 M HClO_4 (Perchloric acid (HClO_4) is a strong acid) [Half-reactions: Ce^+4 + e- Ce^+3 E^degree = +0.771 V; MnO_4^-1 + 5e^- hArr Mn^+2 E^degree = +1.507 V] 5e^- + MnO_4^-1 + 8H^+ = Mn^+2 + 4H_2O Write a balanced net reaction that can occur between the species in this solution. Identify the oxidation and reduction half potential reactions and calculate E

Explanation / Answer

6. pH calculation

a) moles H2CO3 = 0.1 M x 50 ml = 5 mmol

moles NaOH = 0.05 M x 50 ml = 2.5 mmol

HCO3- formed = H2CO3 remained

pH = pKa1 = 6.34

b) moles H2CO3 = 0.1 M x 50 ml = 5 mmol

moles NaOH = 0.15 M x 50 ml = 7.5 mmol

CO3^2- formed = HCO3- remained

pH = pKa2 = 10.36

c) moles H2CO3 = 0.1 M x 50 ml = 5 mmol

moles NaOH = 0.1 M x 50 ml = 5 mmol

HCO3^- formed = 5 mmol/100 ml = 0.05 M

HCO3- + H2O <==> H2CO3 + OH-

Kb2 = 1 x 10^-14/4.7 x 10^-7 = x^2/0.05

x = [OH-] = 3.26 x 10^-5 M

pOH = -log[OH-] = 4.48

pH = 14 - pOH = 9.51

d) moles H2CO3 = 0.1 M x 50 ml = 5 mmol

moles NaOH = 0.2 M x 50 ml = 10 mmol

CO3^2- formed = 5 mmol/100 ml = 0.05 M

CO3^2- + H2O <==> HCO3- + OH-

Kb1 = 1 x 10^-14/4.4 x 10^-11 = x^2/0.05

x = [OH-] = 3.37 x 10^-3 M

pOH = -log[OH-] = 2.47

pH = 14 - pOH = 11.53

7. For the given cell

a) net reaction,

MnO4- + 5Ce3+ + 8H+ --> Mn2+ + 5Ce4+ + 4H2O

oxidation half cell,

Ce3+ --> Ce4+ + e-

reduction half-cell,

MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O

b) dGo = -nFEo

            = -5 x 96500 (1.507 - 0.771) = -355.12 kJ

dGo = -RTlnK

-355120 = -8.314 x 298 lnK

K = 1.78 x 10^62

c) E = Eo - 0.0592/n logK

       = 0.736 - 0.0592/5 [(0.1 x 10^-4/0.1)^6]

       = 1.020 V

d) dG = dGo + RTlnK

          = -355120 + 8,314 x 298 ln[(0.1 x 10^-4/0.1)^6]

          = -492.035 kJ

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