Question

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8. A What is the pH of a solution prepared by dissolving 2.25 mol of chloroacetic acid, (HC2H2O2Cl) in 1.50 L of water? For chloroacetic acid, K. = 1.4 x 10. A) 1.69 B) 2.27 C) 11.7 D) 1.34 E) 2.55

Answer 1

Solution : 8:- Given Datar i) Moles of chloroacetic acid = 2.25 mol ii) Volume of water = 1.5OL m) For chloroacetic acid ka = 1.4x103 > To calcalate - pH of the solution = ? & solution - acetic acid olarity of chloro- - Moles of chloroacetic acid Volume of solution (4) c = 2.25 mol 1.50 L C = 1.5 Molar - For the calculation of pH we need to determine the to lon concentration in the solution, which can be determined by with the help of ka and molarity of chloro acetic acid as followst- 1.4 xlo 3 115 As we know that Degree of dissociation (&) = /ka . a = / 1.400 x = 19.33x104 Also, CH+] = ac =/9.33x104 x 1.5 = 0.0305x1.5 = 0.0458 molar pH = -log[ht] pH = -log[0.0458] pH = 1.34 Answers The correct option is b) pH= 1.34