Question

A sample of carbonic acid (0.125 L, 0.199 M, pK_a1 = 6.35, pK_a2 = 10.33) was titrated with 1.75 M NaOH. Calculate the pH at the following points:

Before the titration.

At the 1st midpoint.

Answer 1

(1) Before titration pH of carbonic acid

First, write out the equilibrium expression, and use the ICE method:

H2CO3 + H2O <---> H3O+ + HCO3-
Initial: 0.199 M 0 0
Change: -x +x +x
Equilibrium: (0.199 - x) x x

Next, set up the Ka expression, and using your given value of Ka1, solve for x.

Ka = [H3O+][HCO3-] / [H2CO3] = 4.3 x 10-7

Water is not included in the expression because it is a liquid. Remember when solving for an equilibrium expression, concentrations of products go in the numerator, and concentrations of reactants go in the denominator.

Ka1 = x2 / (0.199-x) = 4.3 x 10-7

Assume that x is a small, and therefore negligible amount compared to 0.18.

Ka1 = x2 / 0.18 = 4.3 x 10-7

Rearrange to solve for x:

x = [(4.3 x 10-7)(0.199)]1/2 = 2.78 x 10-4 M = [H3O+]

To do this on a calculator: Multiply 4.3 x 10-7 by 0.18, then press the square root key on your calculator.

Now that you have the concentration of the H3O+ ion, you can find the pH of the H2CO3 solution. Remember:

pH = -log[H3O+]

Plug in the concentration you obtained:

pH of H2CO3 solution = -log[2.78 x 10-4 M] = 3.56

(b) At the 1st midpoint

You started out with 125 mL (0.125 L) of 0.199 M H2CO3.

moles H2CO3 = M H2CO3 x L H2CO3 = (0.125)(0.199) = 0.025 moles H2CO3.

Now M1V1 = M2V2

so V2 = (M1V1)/M2

    V2 = (0.199 x 125 ) / 1.75

    V2 = 14.21 ml NaOH

After titrating the first proton with NaOH, you have 0.025 moles of HCO3-. Then after titrating the 2nd proton, you have 0.025 moles of CO3 2-. The total volume of solution at that point = mL sample + mL NaOH = 125 mL + 14.21 mL = 139 mL.

M CO3 2- = moles CO3 2- / L = 0.025 / 0.139 = 0.18 M CO3 2-

We know that CO3 2- is a weak base so it hydrolyzes in water to form HCO3- and OH-.

Molarity . . . . . . .CO3 2- + H2O <==> HCO3- + OH-
Initial . . . . . . .. .0.18 . . . . . . . . .. . . . . . .0 . . . . . .0
Change . . . . . . . . .-x . . . . . . . . . . . . . . . . x . . . . . .x
Equilibrium . . . .0.18-x . . . . . . . . . . . . . . .x . . . . . .x

We need Kb for CO3 2-. Since pKa for its conjugate acid, HCO3-, is 10.33, then
pKb = 14.00 - 10.33 = 3.67 and Kb = 10^-3.67 = 2.1 x 10^-4

Kb = [HCO3-][OH-] / [CO3 2-] = (x)(x) / (0.18-x) = 2.1 x 10^-4
x will be small compared to 0.092 so it can be deleted to simplify the math.
x^2 / 0.18 = 0.00021
x^2 = 3.7 x 10^-5
x = 6.0 x 10^-3 M = [OH-]

pOH = -log [OH-] = 2.22
pH = 14.00 - pOH = 14.00 - 2.22 = 11.77 at he first point