Show all work and answer the following using the correct number of significant f
ID: 229230 • Letter: S
Question
Show all work and answer the following using the correct number of significant figures. [Note: Since this is a take-home quiz and aren't limited by class time, if you need to use the quadratic equation, I expect you do actually do the calculation.] (10 pts) Carbonic acid (H2CO3, Ka 4.46 x 10-7, Ka2 4.69 x 10-11) is a diprotic acid. A titration is performed when 25.00 mL of 0.1250 M disodium carbonate (Na2CO3) is reacted with 0.1535 M HCI. What is the pH at the following points along the titration curve? [Attach a separate sheet if necessary.] a) The initial pH b) 5.00 mL before the first equivalence point c) At the first equivalence point d) 5.00 mL after the first equivalence point e) At the second equivalence point f) 1.00 mL after the second equivalence pointExplanation / Answer
CO3^2- + H+ --> HCO3- ...... Kb1 = 1 x 10^-14/4.69 x 10^-11 = 2.13 x 10^-4
HCO3- + H+ ---> H2CO3 ..... Kb2 = 1 x 10^-14/4.46 x 10^-7 = 2.24 x 10^-8
pKa = -log[Ka]
a) initial pH
Na2CO3 --> 2Na+ + CO3^2-
CO3^2- + H+ --> HCO3- ...... Kb1 = 2.13 x 10^-4
let x amount of CO3^2- hydrolyzed
Kb1 = [HCO3-][OH-]/[CO3^2-]
2.13 x 10^-4 = x^2/0.1250
x = [OH-] = 5.16 x 10^-3 M
pOH = -log[OH-] = 2.29
pH = 14 - pOH = 11.71
b) 5 ml HCl before Ist equivalence point
Volume HCl needed to reach Ist equivalence point = 0.1250 m x 25 ml/0.1535 M = 20.36 ml
5 ml less = 20.36 - 5 = 15.36 ml HCl added
mmol of HCl = 0.1535 M x 15.36 ml = 2.36 mmol
initial moles Na2CO3 present = 0.1250 M x 25 ml = 3.125 mmol
HCO3- formed = 2.36 mmol
CO3^2- remained = 3.125 - 2.36 = 0.765 mmol
using Hendersen-Hasselbalck equation,
pH = pKa + log(base/acid)
= 10.33 + log(0.765/2.36)
= 9.84
c) Ist equivalence point
pH = 1/2(pKa1 + pKa2)
= 1/2(6.35 + 10.33)
= 8.34
d) 5 ml after Ist equivalence point
Volume of HCl added = 20.36 + 5 = 25.36 ml
initial moles Na2CO3 present = 0.1250 M x 25 ml = 3.125 mmol
mmol of HCl = 0.1535 M x 25.36 ml = 3.893 mmol
HCO3- remained = 3.125 - 0.768 = 2.357 mmol
H2CO3 formed = 0.768 mmol
using Hendersen-Hasselbalck equation,
pH = pKa + log(base/acid)
= 6.35 + log(2.357/0.768)
= 6.84
e) IInd equivalence point
[H2CO3] formed = 0.1250 M x 25 ml/40.72 ml = 0.077 M
H2CO3 + H2O <==> H3O+ + HCO3-
let x amount of H2CO3 dissociated
Ka1 = [H3O+][HCO3-]/[H2CO3] = 4.46 x 10^-7
4.46 x 10^-7 = x^2/0.077
x = [H3O+] = 1.85 x 10^-4 M
pH = -log[H3O+] = 3.73
f) 1.00 ml after second equivalence point
excess [H+] = 0.1535 M x 1 ml/41.74 ml = 0.0037 M
pH = -log[H+] = 2.43