Question

The pH of an aqueous solution of 0.335 M nitrous acid is

Answer 1

1. ka for HNO2 = 7128 10-4 ICE Table concentration (M) HNO₂ lag! H+ (aq) + NO₃- (aq) 0.335 -x +00 0,335-x At equilibrium ka = [H +} [NO2- ] 7.2810-4 = xxx (0.335-x) [HNU, As value of ka is very small, so we can neglect x in comparison to 0.335 7.24 810-4 = 002 > x= 1.56 X10-2 M 0.335 = 1.81 [H+] = x= 1.56 X 10 -2 M bH = -log(H+) = -log (1.56 X 10-2) I PH= 1.81 answer

ka for HF = 6.6 810-4 ICE Table concentration (M) HF lag) 0.335 He Cagl & f-lag) I go to tx 0.335-x СС Kaz [1H+] [-] s Kaz xxx = 6.6 A 10-4 [HA] 0.315-1 As value of ko is very small , so we can neglect x in comparison to 0.335 x2 = 6.6810-4 x = 1.49 X 10-2M 0.335 1.83 [ht] = x= 1.498 10-2 M pH = -log (H+] = -log (1149 X 10-2) = port- 14-pH = 14-1183 = 12.17 12.17 po H = Port = 12.17 Arower