Question

4) 5.65 mL of 0.23 M Ba(OH)2 is delivered while titrating of 28.0 mL of 0.390 M HCN. The volumes are additive and the temperature of the solution is 25°C. Ka for hydrocyanic acid is 6.2 x 100. a. How many moles of OH were added? b. Find [H] in the final solution. (Hint: You could find pH first, then [H'] from pH.) C. Find [OH] in the final solution. (Hint: Use Kw)

Answer 1

giver, Concentration of Bacon) = 0.23M volume of Ba(OH), sol = 5.65m2 = 5.65x18% moles of BacoH)2 = concentration x volume (192) = 0.23M X 5.65x163L = 1.3x10 mol BaloH) is a strong base it is dissociated into Bat2 and 2 on eon.. 30 moles of ort ion = 1.3x108x2 = 2.6x103 moh a) 2.6x103 moles of Oh ion added. An) concentration of HCN soi = 0.3909 volume & HeN sol = 28.0m = 28.0x1031 moles of How = 0.390M X 28.08103 L = 109 x162 mol

H₂O + (N- 0 HON + OH Intial 1.09810² 1.3x103 change 1.3x10 3 -1.3x103 13x10 at ezen (9.62x17%) 0 - 1 38103 concentration at equilibrium (fi tinal volume of soi = 98.0 x10314 5.65x103 L - 33.65 x103 L [HeN] = 9.65 x 10 mot _ 0.2868m 33.65x103 [cat] = 1.3x103 mol - != 0.0386 M 33-65X103 → To calculate the pH of the buffer sol P4 = pkę ([N ] CHINI p4 = pka + rog (Text) fos-rugha puz tog (-23 15")-13 (34)

PH = -log (4) 8.34 = -log (+) [H ] = 108.34 = 4.57X100 g (Ams) : Kw = [H'][on] 1014 = (4.57X109) Corr] [06] = 1014 4.57 xias = 2.188x106m (ans).