Question

Consider the titration of a 40.0 ml. of 0.155 M weak acid HA (Ka = 2.7 x 10") with 0.100 M LiOH. What is the pH of the solution before any base has been added? L 4 points b What would be the pH of the solution after the addition of 200 ml of LiOH? 4 points How many mL of the LiOH would be required to reach the halfway point of the titration? 1 4 points points d What is the pH of the solution at the equivalence point? 4 points What would be the pH of the solution after that addition of 100.0 mL of LiOH? 4 points

Answer 1

a)

Reaction	HA(aq)	$\rightleftharpoons$	H+(aq)	+	A-(aq)
Initial Conc.	0.155 M		0 M		0 M
Change in Conc.	-y M		+y M		+y M
Equilibrium Conc.	(0.155 - y) M		y M		y M

Ka = [H+][A-]/[HA] = 2.7 x 10-8

(y)(y)/(0.155 - y) = 2.7 x 10-8

y2 + (2.7 x 10-8)y - (2.7 x 10-8 x 0.155) = 0

y = 6.47 x 10-5 M

or [H+] = 6.47 x 10-5 M

So, pH = -log10([H+]) = -log10(6.47 x 10-5) = 4.19

Therefore, pH = 4.19 before adding any base to the solution.

b) Concentration of HA = 0.155 M = 0.155 mol/L

Volume of HA = 40.0 mL = (40.0 mL)(1 L/1000 mL) = 0.0400 L

Moles of HA = (0.155 mol/L)(0.0400 L) = 6.20 x 10-3 mol

Concentration of LiOH = 0.100 M = 0.100 mol/L

Volume of LiOH = 20.0 mL = (20.0 mL)(1 L/1000 mL) = 0.0200 L

Moles of LiOH = (0.100 mol/L)(0.0200 L) = 2.00 x 10-3 mol

Reaction	HA(aq)	+	LiOH(aq)	$\rightarrow$	LiA(aq)	+	H2O(l)
Initial Moles	6.20 x 10-3 mol		2.00 x 10-3 mol		0 mol		PURE
Change in Moles	-2.00 x 10-3 mol		-2.00 x 10-3 mol		+2.00 x 10-3 mol
Final Moles	4.20 x 10-3 mol		0 mol		2.00 x 10-3 mol		LIQUID

Total Volume of Solution = 40.0 mL + 20.0 mL = 60.0 mL = (60.0 mL)(1 L/1000 mL) = 0.0600 L

Concentration of LiA = Concentration of A- = (2.00 x 10-3 mol)/(0.0600 L) = 0.0333 mol/L = 0.03 M

Concentration of HA = (4.20 x 10-3 mol)/(0.0600 L) = 0.07 mol/L = 0.07 M

Now,

Reaction	HA(aq)	$\rightleftharpoons$	H+(aq)	+	A-(aq)
Initial Conc.	0.07 M		0 M		0.03 M
Change in Conc.	-y M		+y M		+y M
Equilibrium Conc.	(0.07 - y) M		y M		(0.03 + y) M

Ka = [H+][A-]/[HA] = 2.7 x 10-8

(y)(0.03 + y)/(0.07 - y) = 2.7 x 10-8

y2 + (0.03 + 2.7 x 10-8)y - (2.7 x 10-8 x 0.07) = 0

y = 5.67 x 10-8 M

or [H+] = 5.67 x 10-8 M

So, pH = -log10([H+]) = -log10(5.67 x 10-8) = 7.25

Therefore, pH = 7.25 after adding 20.0 mL of base to the solution.

c) Concentration of HA = 0.155 M = 0.155 mol/L

Volume of HA = 40.0 mL = (40.0 mL)(1 L/1000 mL) = 0.0400 L

Moles of HA = (0.155 mol/L)(0.0400 L) = 6.20 x 10-3 mol

Concentration of LiOH = 0.100 M = 0.100 mol/L

Volume of LiOH = V mL = (V mL)(1 L/1000 mL) = 0.001V L

Moles of LiOH = (0.100 mol/L)(0.001V L) = 1.00V x 10-4 mol

Reaction	HA(aq)	+	LiOH(aq)	$\rightarrow$	LiA(aq)	+	H2O(l)
Initial Moles	6.20 x 10-3 mol		1.00V x 10-4 mol		0 mol		PURE
Change in Moles	-1.00V x 10-4 mol		-1.00V x 10-4 mol		+1.00V x 10-4 mol
Final Moles	(6.20 x 10-3 - 1.00V x 10-4) mol		0 mol		1.00V x 10-4 mol		LIQUID

At Half Equivalence Point,

Moles of LiOH = (1/2)(Moles of HA)

Moles of LiOH = 1.00V x 10-4 = (1/2) 6.2 x 10-3 = 3.1 x 10-3

So, V = 3.1 x 10-3/10-4 mL = 31.0 mL

Therefore, 31.0 mL of LiOH is needed to half-equivalence point.

d) Concentration of HA = 0.155 M = 0.155 mol/L

Volume of HA = 40.0 mL = (40.0 mL)(1 L/1000 mL) = 0.0400 L

Moles of HA = (0.155 mol/L)(0.0400 L) = 6.20 x 10-3 mol

Concentration of LiOH = 0.100 M = 0.100 mol/L

Volume of LiOH = V mL = (V mL)(1 L/1000 mL) = 0.001V L

Moles of LiOH = (0.100 mol/L)(0.001V L) = 1.00 x 10-4V mol

Reaction	HA(aq)	+	LiOH(aq)	$\rightarrow$	LiA(aq)	+	H2O(l)
Initial Moles	6.20 x 10-3 mol		1.00V x 10-4 mol		0 mol		PURE
Change in Moles	-1.00V x 10-4 mol		-1.00V x 10-4 mol		+1.00V x 10-4 mol
Final Moles	(6.20 x 10-3 - 1.00V x 10-4) mol		0 mol		1.00V x 10-4 mol		LIQUID

At Equivalence Point,

Moles of LiOH = Moles of HA

Moles of LiOH = 1.00V x 10-4 = 6.2 x 10-3

So, V = 6.2 x 10-3/10-4 mL = 62.0 mL

Total Volume of Solution = 40.0 mL + 62.0 mL = 102.0 mL = (102.0 mL)(1 L/1000 mL) = 0.102 L

Concentration of LiA = Concentration of A- = (6.20 x 10-3 mol)/(0.102 L) = 0.0608 mol/L = 0.0608 M

Concentration of HA = (0 mol)/(0.102 L) = 0 mol/L = 0 M

Now,

Reaction	A-(aq)	+	H2O(aq)	$\rightarrow$	HA(aq)	+	OH-(aq)
Initial Conc.	0.0608 M		PURE		0 M		0 M
Change in Conc.	-a M				+a M		+a M
Final Conc.	(0.0608 - a) M		LIQUID		a M		a M

Keq = Kw/Ka = [HA][OH-]/[A-] = (10-14)/(2.7 x 10-8)

a2/(0.0608 - a) = (10-14)/(2.7 x 10-8)

a2 + (3.7 x 10-7)a - (3.7 x 10-7 x 0.0608) = 0

a = 1.49 x 10-4 M

or [OH-] = 1.49 x 10-4 M

So, pOH = -log10([OH-]) = -log10(1.49 x 10-4) = 3.82

pH = 14 - pOH = 14 - 3.82 = 10.17 $\approx$ 10.2

Therefore, pH = 10.2 at equivalence point.

e) Concentration of HA = 0.155 M = 0.155 mol/L

Volume of HA = 40.0 mL = (40.0 mL)(1 L/1000 mL) = 0.0400 L

Moles of HA = (0.155 mol/L)(0.0400 L) = 6.20 x 10-3 mol

Concentration of LiOH = 0.100 M = 0.100 mol/L

Volume of LiOH = 100.0 mL = (100.0 mL)(1 L/1000 mL) = 0.100 L

Moles of LiOH = (0.100 mol/L)(0.100 L) = 1.00 x 10-2 mol

Reaction	HA(aq)	+	LiOH(aq)	$\rightarrow$	LiA(aq)	+	H2O(l)
Initial Moles	6.20 x 10-3 mol		1.00 x 10-2 mol		0 mol		PURE
Change in Moles	-6.20 x 10-3 mol		-6.20 x 10-3 mol		+6.20 x 10-3 mol
Final Moles	0 mol		3.80 x 10-3 mol		6.20 x 10-3 mol		LIQUID

Total Volume of Solution = 40.0 mL + 100.0 mL = 140.0 mL = (140.0 mL)(1 L/1000 mL) = 0.140 L

Concentration of LiA = Concentration of A- = (6.20 x 10-3 mol)/(0.140 L) = 0. mol/L = 0.0443 M

Concentration of LiOH = Concentration of OH- = (3.80 x 10-3 mol)/(0.140 L) = 0.0271 mol/L = 0.0271 M

Now,

Reaction	A-(aq)	+	H2O(aq)	$\rightarrow$	HA(aq)	+	OH-(aq)
Initial Conc.	0.0443 M		PURE		0 M		0.0271 M
Change in Conc.	-a M				+a M		+a M
Final Conc.	(0.0443 - a) M		LIQUID		a M		(0.0271 + a) M

Keq = Kw/Ka = [HA][OH-]/[A-] = (10-14)/(2.7 x 10-8)

a(0.0271 + a)/(0.0443 - a) = (10-14)/(2.7 x 10-8)

a2 + (0.0271 + 3.7 x 10-7)a - (3.7 x 10-7 x 0.0443) = 0

a = 0.197 M

or [OH-] = 0.197 M

So, pOH = -log10([OH-]) = -log10(0.197) = 0.70

pH = 14 - pOH = 14 - 0.7 = 13.3

Therefore, pH = 13.3 after addition of 100.0 mL of base to the solution.