Question

A 30.0- mL sample of 0.165 M propanoic acid is titrated with 0.300 MKOH . Part A Calculate the pH at 0 mL of added base. Part B Calculate the pH at 5 mL of added base. Part C Calculate the pH at 10 mL of added base. Part D Calculate the pH at the equivalence point. Part E Calculate the pH at one-half of the equivalence point. Part F Calculate the pH at 20 mL of added base. Part G Calculate the pH at 25 mL of added base.

Answer 1

pK_a of propanoic acid (CH₃CH₂COOH) is 4.87. Represent propanoic acid as HA.

Part A

Consider the dissociation of Haas below:

HA (aq) <====> H⁺ (aq) + A^-(aq)

initial                                    0.165                   0              0

change                                  -x                       +x             +x

equilibrium                     (0.165 – x)               x               x

Acid dissociation constant: K_a = [H⁺][A^-]/[HA] = (x).(x)/(0.165 – x)

Given pK_a = 4.87, K_a = 10^-4.87 = 1.349*10^-5. Therefore,

K_a = x²/(0.165 – x)

====> 1.349*10^-5 = x2/(0.165 – x)

Make an assumption: x is much smaller than 0.165 so that (0.165 – x) ≈ 0.165; therefore,

1.349*10^-5 = x²/0.165

===> x² = 2.22585*10^-6

===> x = 1.492*10^-3

Therefore, [H⁺] = 1.492*10^-3 M and pH = - log [H⁺] = -log (1.492*10^-3) = 2.826 ≈ 2.83 (ans).

Part B

Moles HA added = (volume of HA added in L)*(concentration in mol/L) = (30.0 mL)*(1 L/1000 mL)*(0.165 mol/L) = 0.00495 mol.

Moles KOH added = (volume of KOH added in L)*(concentration of KOH in mol/L) = (5.00 mL)*(1 L/1000 mL)*(0.300 mol/L) = 0.0015 mol.

Set up the ICE chart for the reaction as below:

HA (aq) + KOH (aq) ------> KA (aq) + H₂O (aq)

initial                      0.00495   0.0015                     0

change                    -0.0015   -0.0015                +0.0015

equilibrium             0.00345        0                       0.0015

Total volume of the solution = (30.00 + 5.00) mL = 35.00 mL.

[HA]_e = (0.00345/35.00) mM

[KA]_e = (0.0015/35.00) mM

Use the Henderson-Hasslebach equation to find the pH as

pH = pK_a + log [KA]_e/[HA]_e = 4.87 + log [(0.0015/35.00)/(0.00345/35.00)] = 4.87 + log (0.0015/0.00345) = 4.87 + log (0.43478) = 4.508 ≈ 4.51 (ans).

Part C

Moles of KOH added = (10.00 mL)*(1 L/1000 mL)*(0.300 mol/L) = 0.003 mol.

Total volume of solution = (30.00 + 10.00) mL = 40.00 mL.

Moles of HA at equilibrium = (0.00345 – 0.003) = 0.00045 mol.

Moles of KA at equilibrium = 0.003 mol (we can find out these two values by putting up an ICE chart as above).

[HA]_e = (0.00045/40.00) mM

[KA]_e = (0.003/40.00) mM

Use the Henderson-Hasslebach equation:

pH = pK_a + log [KA]_e/[HA]_e = 4.87 + log [(0.003/40.00)/(0.00045/40.00)] = 4.87 + log (0.003/0.00045) = 5.6939 ≈ 5.69 (ans).

Part D

Calculate the volume of KOH required for complete neutralization of HA with KOH using the dilution law as

(30.00 mL)*(0.165 M) = V*(0.300 M)

====> V = 16.5 mL

Volume of KOH required for titration = 16.5 mL.

Moles of KOH consumed at the equivalence point = moles of KA produced = (16.5 mL)*(1 L/1000 mL)*(0.300 mol/L) = 0.00495 mol.

KA is the conjugate base of the weak acid and re-establishes equilibrium as

KA (aq) + H₂O (l) <====> HA (aq) + KOH (aq)

Since KOH (base is produced), we must work with the pK_b of KA. Given pK_a = 4.87,

pK_b = 14 – pK_a = 14 – 4.87 = 9.13

===> K_b = 10^-9.13 = 7.41*10^-10

Total volume of the solution = (30.00 + 16.5) mL = 46.5 mL = 0.0465 L.

[KA] = (0.00495/0.0465) mol/L = 0.106 M

We must have

K_b = [HA][KOH]/[KA] = x²/(0.106 – x)

Small x approximation:

7.41*10^-10 = x²/0.106

====> x² = 7.8546*10^-11

====> x = 8.8626*10^-6 ≈ 8.863*10^-6

Therefore, [KOH] = 8.863*10^-6 M and pOH = -log [OH^-] = -log(8.863*10^-6) = 5.052

pH = 14 – pOH = 14 – 5.052 = 8.948 ≈ 8.95 (ans).