(a) (i) Calculate the [H3O+] and pH of a solution containing 1.38 × 10-2 mol L-1

Question

(a) (i) Calculate the [H3O+] and pH of a solution containing 1.38 × 10-2 mol L-1 bromoacetic acid (pKa = 2.902) and sufficient potassium chloride to make I = 0.10 mol L-1. Use activities in this calculation.

(ii) Calculate the [H3O+] and pH at the equivalence point of a titration of 20.00 mL of 2.45 × 10-2 mol L-1 ethylamine (CH3CH2NH2) with 0.09982 mol L-1 HCl. Ignore activities in this calculation. pKa(CH3CH2NH3 + = 10.673)

(b) A 25.00 mL volume of a solution containing 0.01966 mol L-1 of the diprotic acid malonic acid (pKa1 = 2.847, pKa2 = 5.696) is titrated with 0.04836 mol L-1 NaOH. Ignore activities for the following calculations.

(i) Calculate the [H3O+] and pH after the addition of 7.00 mL of NaOH.

(ii) Calculate the [H3O+] and pH after the addition of 15.00 mL of NaOH.

(iii) Calculate the [H3O+] and pH after the addition of 25.00 mL of NaOH.

(c) (i) A 10.00 mL volume of 0.02762 mol L-1 Ca2+ solution is added to 50.00 mL of pH 9 buffer and the resulting solution is titrated with 0.04971 mol L-1 EDTA solution. Calculate the volume of EDTA added at the equivalence point and the calcium ion concentration at the equivalence point, ignoring activities. logKf(CaEDTA2-) = 10.79 (EDTA4-) = 5.4 × 10-2 at pH 9

(ii) A calcium-containing sample with mass 1.836 g is dissolved in water and the volume is made up to 100.0 mL. A 20.00 mL aliquot of this solution is placed in a flask, 50.00 mL of pH 9 buffer and a small amount of solid indicator are added, and the resulting solution is titrated with 0.04971 mol L-1 EDTA solution to an endpoint at 18.03 mL. Calculate the mass % of Ca in the sample. (Molar mass for Ca = 40.078 g mol-1)

Explanation / Answer

(a) (i) [acid] = 1.38 X 10-2 mol L-1             [salt] = 0.10 mol L-1

pKa = 2.902

pH = pKa + log ([salt]/[acid])

pH = 2.902 + log(0.1/1.38 X 10-2) = 3.76

pH = -log[H3O]+ = 3.76

[H3O]+ = antilog(-3.76) = 1.74 X 10-4 mol L-1

(ii) At the equivalence point V1 X N1 = V2 X N2

Volume of acid, V1 = 20 X 2.45 X 10-2/0.09982 = 4.91 mL

pKa = 10.673

Therefore Ka = antilog(-10.673) = 4.71 X 10-10 mol L-1

[CH3CH2NH3+] at the end point = volume of CH3CH2NH3+/total volume = 2.45 X 10-2/0.02491 = 1 M

4.71 X 10-10 = X2/1 +x

This is a quadratic equation and solving this we get x = 2.17 X 10-5 mol L-1 = [H3O+]

pH = -log[H3O+] = 4.66

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