Question

What is the equilibrium pH of an initially 0.700M for a monoprotic acid with a K_a = 7.687E-4

Answer 1

I have provided the solution. Just go through it.
I Let the monoprofic acid be-HX 80 et dissociates as HX -H + + x Tretially 0 0 1 0 out 0 0 at fro at of = teq (0.7-2M RM RM (at e quelli - acid .) breium so dissociation constant ka can be given here as :- 100 ka = x x x - (017-2)

XH geven 7 ka = 7.688 1-4 or tota 768X1042 07-27 ² (7.68 x 10) (0.7) 2 (7.687 104 ) * ** + 4.68x10x = ) 53.46X10% =0 2 -768x102V G. 68 x 1859) (3-6 4x(7.68x109 - 5.376810 810.0=1 otsait 2X1 = -7.68X10° + 58.987 1084 8. 165.15 8108 = -7.68x109114.97X169

P = 3.64x10^M = [H] a pH = -log CH] = -109 (3.64x104) pH = 3.43 AM