Question

30.0 mL of 0.100 M H2CO3 is titrated with 0.200 M KOH. Calculate the initial pH before KOH has been added. ka = 4.3 x 10-7. Calculate the pH when 10.0 mL of a .200M KOH is added to 30.0 mL of 0.100 M H2CO3. ka = 4.3 x 10-7. Calculate the equivalence point and then calculate the pH at the equivalence point. Calculate the pH if 20.0 mL of 0.200 M KOH is added to 30.0 mL of 0.100 M H2CO3.

Answer 1

Calculation of initial pH before KOH has been added:

                                                          H₂CO₃                          H⁺               +              HCO₃^-

Initial concentration                            0.100 M                             0                                   0

Change in concentration                       - X                                   X                                   X

Equilibrium concentration              (0.100 - X)                              X                                   X

Now at equilibrium,

Ka = [H⁺][HCO₃^-]/[H₂CO₃]

or, Ka = X²/(0.100 - X)

or, 4.3 x 10^-7 = X²/(0.100 - X)

or, 4.3 x 10^-8 - 4.3 x 10^-7X = X²

or, X² + 4.3 x 10^-7X - 4.3 x 10^-8 = 0

or,

or

or,

X = 0.00041457/2 (Taking the positive value)

   = 0.000207285

X = [H+] = 0.000207285

Therefore,

pH = -log[H+] = - log0.000207285 = 3.68

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Calculation the pH when 10.0 mL of 0.200M KOH is added

H₂CO₃ + KOH KHCO₃ + H₂O

10.0 mL of 0.200M KOH = 10 x 0.200 = 2 mmol of KOH

30.0 mL of 0.100 M H₂CO₃ = 30 x 0.100 = 3 mmol of H₂CO₃

2 mmol of KOH will produce 2 mmol of KHCO₃

Remaining H₂CO₃ = (3-2) = 1 mmol

Total volume of the solution = (10 + 30) = 40 mL

Molarity of KHCO₃ = 2/40 = 0.05 M

Molarity of H₂CO₃ = 1/40 = 0.025 M

From Henderson-Hasselbach equation,

pH = pK_a + log[KHCO₃][H2CO₃]

      = - log[4.3 x 10^-7] + log 0.05/0.025

      = 6.36 + 0.30

      = 6.66

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Calculation of equivalence point and pH at equivalence point

Concentration of KOH, C₁ = 0.200 M

Volume of KOH, V₁ = ?

Concentration of H₂CO₃, C₂ = 0.100 M

Volume of H₂CO₃, V₂ = 30 mL

Now,

V₁C₁ = V₂C₂

or V₁ = (30 x 0.100)/0.200 = 15 mL

Therefore the volume of KOH required to reach the equivalence point = 15 mL

Moles of KHCO₃ = 15 x 0.200 = 3 mmol

Molarity of KHCO₃ = 3/ (15 + 30) = 0.07 M

At this point all of H₂CO₃ will convert to KHCO₃ and KHCO₃ will hydrolyze to produce hydroxide ions.

                                                             HCO₃^- + H₂O H₂CO₃ + OH^-

Initial concentration                             0.07                        0              0

Change in concentration                     -X                           X              X

Equilibrium concentration               (0.07 - X)                    X              X

K_a = 4.3 x 10-7

pK_a = 6.36

pK_b = pK_w - pK_a = 14 - 6.36 = 7.64

or, Kb = 10^-7.64 = 2.29 x 10^-8

Kb = [H₂CO₃][OH-]/[HCO₃^-]

or, 2.29 x 10^-8 = X²/(0.07 - X) = X²/0.07    (We can assume (0.07- X) = 0.07 as X is very small)

or, X² = 1.60 x 10^-9

or, X = 4.00 x 10^-5

Thus X = pOH = -log[OH^-] = -log 4.00 x 10^-5 = 4.40

Now, pH = 14 - pOH = 14 - 4.40 = 9.60

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Calculation of pH if 20.0 mL of 0.200 M KOH is added

Moles of KOH = 20 x 0.200 = 4 mmol

Moles of H₂CO₃ = 30 x 0.100 = 3 mmol

After reaction with H2CO3 the moles of KOH will remain = 1 mmol

Total volume of the solution = (30 + 20) = 50 mL

Molarity of remaining KOH = 1/50 = 0.02

pOH = -log0.02 = 1.70

Therefore now,

pH = 14 - pOH = 14 - 1.70 = 12.3