Consider a 26.0 mL aliquot of a 0.1180 M aqueous solution of the weak base (B),
ID: 513457 • Letter: C
Question
Consider a 26.0 mL aliquot of a 0.1180 M aqueous solution of the weak base (B), ethyl 3-aminobenzoate, characterized by a base-dissociation constant, Kb = 4.96E–8. This solution is titrated with 0.0772 M HCl. Calculate or determine the following (work must be shown for credit; to simplify grading, please use the “small x” approximation – it is valid here):
(a) The “initial pH” (i.e. the pH of the analyte solution):
(b) The volume (in mL) of titrant needed to reach the endpoint: .
(c) The pH at the “equivalence point”:
(d) The pH after the addition of 23.5 mL of titrant: .
(e) The pH after the addition of 42.0 mL of titrant: .
(f) The pH after the addition of a “large excess” of titrant: .
9.88/39.74mL/4.01/6.54/2.59/1.11 (these are the answers I just dont know how to get them)
Explanation / Answer
a) Consider the reaction of the base with water as
B (aq) + H2O (l) -------> BH+ (aq) + OH- (aq)
Kb = [BH+][OH-]/[B] = (x).(x)/(0.1180 – x) (see the 1:1 dissociation)
We use the small x approximation and write
4.96*10-8 = x2/0.1180
===> x2 = 5.8528*10-9
===> x = 7.650*10-5
Therefore, [BH+] = [OH-] = 7.650*10-5 M and pOH = -log [OH-] = -log (7.650*10-5) = 4.116
We know that pH + pOH = 14; therefore, pH = 14 – pOH = 14 – 4.116 = 9.884 9.88 (ans).
b) Write down the neutralization reaction between B and HCl as below:
B (aq) + HCl (aq) ------> BH+ (aq) + Cl- (aq)
As per the stoichiometric balanced equation above, we must have
1 mole B = 1 mole HCl
Moles of B added = (26.0 mL)*(0.1180 mol/L) = 3.068 mmole.
Therefore, moles of HCl required for titration = 3.086 mmole.
Concentration of the supplied HCl solution = 0.0772 M.
Therefore, volume of HCl required = (3.086 mmole)/(0.0772 mol/L) = 39.741 mL 39.74 mL (ans).
c) At the equivalence point, we have B completely neutralized and BH+. BH+ is the conjugate acid of the weak base, B and establishes equilibrium as
BH+ + H2O -----> B + H3O+
The acid ionization constant Ka is given as
Ka = [H3O+][B]/[BH+]
Given Kb, we have Ka*Kb = Kw where Kw = 1.0*10-14; therefore, Ka = (1.0*10-14)/(4.96*10-8) = 2.016*10-7
Volume of the solution = (26.0 + 39.74) mL = 65.74 mL.
Moles of BH+ formed = moles of B neutralized = 3.086 mmole.
[BH+] = (3.086 mmole)/(65.74 mL) = 0.0469 M
Plug in the expression for Ka.
2.016*10-7 = (x).(x)/(0.0469 – x)
Use the small x approximation and write
2.016*10-7 = x2/0.0469
===> x2 = 9.45504*10-9
===> x = 9.723*10-5
Therefore, [H3O+] = 9.723*10-5 M and pH = -log [H3O+] = -log (9.723*10-5) = 4.012 4.01 (ans).
d) Moles of HCl added = (23.5 mL)*(0.0772 mol/L) = 1.8142 mmole.
Moles of B neutralized = 1.8412 mmole; moles of BH+ formed = 1.8412 mmole.
Moles of B retained = (3.068 – 1.8412) mmole = 1.2268 mmole.
Volume of solution = (26.0 + 23.5) mL = 49.5 mL.
Concentration of BH+ = (1.8412 mmole)/(49.5 mL) = (1.8412/49.5) M
Concentration of B = (1.2268 mmole)/(49.5 mL) = (1.2268/49.5) M
Use the Henderson-Hasslebach equation for weak base and write
pOH = pKb + log [BH+][[B] = -log (4.96*10-8) + log [(1.8412/49.5) M/(1.2268/49.5) M] = 7.3045 + log (1.8412/1.2268) = 7.3045 + 01757 = 7.4802
We know that pH + pOH = 14; therefore, pH = 14 – pOH = 14 – 7.4802 = 6.5198 6.52 (ans).
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