Question 5 We dissolve 3.44 g of an unknown acid, HA, in water in order to produ

Question

Question 5 We dissolve 3.44 g of an unknown acid, HA, in water in order to produce 25.0 mL of solution The pH of this solution is 1.55. We titrate this solution with a solution of NaOH (aq) 0.250 M and reach the equivalent point after having added 18.8 ml of the NaOH 0.250 solution. a) (2 points) What is the molar mass of HA? b) (3 points) What is the value of the dissociation constant of HA (aq)? c) 3 points) During the titration, what would have been the pH after the addition of 13.3 mL of the 0.250 M NaOH solution?

Explanation / Answer

a) First we apply the neutralization relation with the equivalence point, to know the concentration of the acid:

Ca * Va = Cb * Vb

We cleared Ca:

Ca = Cb * Vb / Va = 0.250 M * 18.8 mL / 25 mL = 0.188 M

We determine the moles of acid present in the sample:

0.188 mol / L * 0.025 L = 4.7x10 ^ -3 moles

and we determine molecular weight:

MW = mass / moles = 3.44 gr / 4.7x10 ^ -3 moles = 732 gr / mol.

b) Of the reaction:

HA + H2O = H3O + + A-

we have for pH = 1.55

[H3O +] = Antilog (-pH) = 0.0281 M = [A-]

[HA] = 0.188 - 0.0281 = 0.16 M

we calculate Ka = [H3O +] * [A-] / [HA] = (0.0281 * 0.0281) / 0.16 = 5x10 ^ -3

c) We calculate the aggregate base moles and subtract them from the moles of acid present:

nOH = 0.250 mol / L * 0.0133 L = 3.33x10 ^ -3 moles

nHA remaining = 4.7x10 ^ -3 - 3.33x10 ^ -3 = 1.37 x10 ^ -3 moles remaining

we calculate the final concentration of HA:

[HA] = 1.37 x10 ^ -3 / 0.0383 L = 0.036 M

and the concentration of NaOH in the new volume that would be equal to the A-final concentration:

[NaOH] = [A-] = 3.33x10 ^ -3 / 0.0383 = 0.087 M

and with the following equation we calculate pH:

pH = pKa - Log [HA] / [A-] = 2.3 - Log (0.036 / 0.087) = 2.68.

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