Calculate the pH of a solution prepared by mixing 0.0826 mol of chloroacetic aci

Question

Calculate the pH of a solution prepared by mixing 0.0826 mol of chloroacetic acid plus 0.0413 mol of sodium chloroacetate in 1.00 L of water. (Assume pKa = 2.866 and Kw = 1.011014. Enter unrounded values.)

(a) First do the calculation by assuming that the concentrations of HA and A equal their formal concentrations. 2.564970004 Correct: Your answer is correct.

(b) Then do the calculation using the real values of [HA] and [A] in the solution.

(c) Using first your head, and then the Henderson-Hasselbalch equation, find the pH of a solution prepared by dissolving all the following compounds in one beaker containing a total volume of 1.00 L: 0.174 mol ClCH2CO2H, 0.026 mol ClCH2CO2Na, 0.074 mol HNO3, and 0.074 mol Ca(OH)2. Assume that Ca(OH)2 dissociates completely.

Explanation / Answer

(a) pH = 2.866 + log(0.0413/0.0826) = 2.56497

(b) Ka = 1.36 x 10^-3 = (0.0413 + x)^2/(0.0826 - x)

1.12 x 10^-4 - 1.36 x 10^-3x = 3.41 x 10^-3 + 0.1239x + x^2

x^2 + 0.12254x - 1.12 x 10^-4 = 0

x = [H+] = 8.54 x 10^-3 M

pH = 2.07

(c) pH of mixture

2 mols of HNO3 reacts with 1 mle of Ca(OH)2

0.074 mols HNO3 reacted with 0.037 mols of Ca(OH)2

remaining Ca(OH)2 = 0.037 mols

final moles of ClCH2CO2H = 0.174 - 2 x 0.037 = 0.1mols

final moles of ClCH2CO2Na = 0.026 + 2 x 0.037 = 0.1 mols

[ClCH2CO2H] = 0.1 M

[ClCH2CO2Na] = 0.1 M

pH = pKa = 2.866

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