Question

11. Calculate the pH of a solution of 0.10 M hydrogen cyanide HON? (Ka = 4.9 x 10-10) E. 9.31 F. 5.15 G. 4.16 H. 1.00

Answer 1

as of HeN tan ese written - CN- (aq) + 130+ (ar ta ta 11. Ans; - The ionisation HON (aq) + 40lag) At t-o, 0.10 m At t=teg, (0.10-x) ka= [[N] [ 20t] [HON] a, ka = (a (a) 0.10-2 But given that Ka = 4.9 x 1000 .:ka= 22 0-10-1 a, 4.9.x1810 22 0.10-2

0.10-2 ~ 0.10. Tinco x is small in comparison to 0.10,50 Thus, above equation becomes 4.9 x 100 = x a) 2 = 0.49 x1050 = x= 10.49 * 1810 2) = 0.7 x 10-5 .. [hot] = 0.7x105 m. But [hot] = [4] i pH = -log [Bot] = -log(0.7x105) = -log(0.7) - log (105) =-(-0.154) + 5 = 5+0.159 pH = 5.15 is 5015 Thurs The pH of correct 0.10 m option HeN solution is 05:15