Show all work and answer the following using the correct number of significant f
ID: 229688 • 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, Kal = 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 (Na2C03) 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 = Kw/Ka2 = 1 x 10^-14/4.69 x 10^-11 = 2.13 x 10^-4
HCO3- + H+ <==> H2CO3 ..... Kb2 = Kw/Ka1 = 1 x 10^-14/4.46 x 10^-7 = 2.24 x 10^-8
pKa1 = -log(4.46 x 10^-7) = 6.35
pKa2 = -log(4.69 x 10^-11) = 10.33
a) initial
[CO3^2-] = 0.125 M
hydrolysis reaction,
CO3^2- + H2O <==> HCO3- + OH-
let x amount has hydrolyzed
Kb1 = [HCO3-][OH-]/[CO3^2-]
2.13 x 10^-4 = x^2/0.125
x = [OH-] = 5.16 x 10^-3 M
pOH = -log(5.16 x 10^-3) = 2.29
pH = 14 - 2.29 = 11.71
b) 5 ml before Ist equivalence point
Volume HCl needed to reach Ist equivalence point = 0.125 M x 25 ml/0.1535 M = 20.36 ml
So,
volume HCl added = 15.36 ml
initial moles CO3^2- = 0.125 M x 25 ml = 3.125 mmol
HCl added = 0.1535 M x 15.36 ml = 2.358 mmol
(HCO3-) formed = 2.358 mmol
(CO3^2-) remained = 0.767 mmol
Using Hendersen-Hasselbalck equation,
pH = pKa2 + log(CO3^2-/HCO3-)
= 10.33 + log(0.767/2.358)
= 9.84
c) at Ist equivalence point
all of CO3^2- has reacted to form HCO3-
pH = 1/2(pKa1 + pKa2)
= 1/2(6.35 + 10.33)
= 8.34
d) 5 ml after Ist equivalence point
Volume HCl needed to reach Ist equivalence point = 0.125 M x 25 ml/0.1535 M = 20.36 ml
So,
volume HCl added = 25.36 ml
initial moles CO3^2- = 0.125 M x 25 ml = 3.125 mmol
HCl added = 0.1535 M x 25.36 ml = 3.893 mmol
(H2CO3) formed = 0.767 mmol
(HCO3-) remained = 2.358 mmol
Using Hendersen-Hasselbalck equation,
pH = pKa2 + log(CO3^2-/HCO3-)
= 6.35 + log(2.358/0.767)
= 6.84
e) at IInd equivalence point
all of HCO3- has reacted to form H2CO3
[H2CO3] formed = 0.125 M x 25 ml/(2 x 20.36 ml) = 0.077 M
Dissociation of weak acid
H2CO3 + H2O <==> H3O+ + HCO3-
let x amount dissociated
Ka1 = [HCO3-][H3O+]/[H2CO3]
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 IInd equivalence point
excess [H+] = 0.1535 M x 1.0 ml/41.72 ml = 0.0037 M
pH = -log[H+] = -log(0.0037) = 2.43