A 0.68 M solution of a weak acid with general formula HX is found to have a pH of 6.3

Question

a) Finish the reaction when acid is added to water HX(aq) + H₂O(I)

b) Write an expression for the acid equilibrium constant Ka

c) Find [H₃O⁺]

d) Calculate the value of Ka for this acid

Answer 1

a) HX + H2O ==> H₃O⁺ + X^-

b) Ka = [H3O⁺][X^-] / [HX]

c) pH = 6.3 and pH = -log [H3O⁺]

[H3O⁺] = 1x10^-6.3 = 5.0x10^-7

d) Ka = (5.0x10^-7)(5.0x10^-7) / 0.68

Ka = 3.68x10^-13

Answer 2

An acid gives a hydrogen ion to something else, so in this example... HX(aq) + H₂O(I) ->X- + H30+.

The eqilibium constant is simply products over reactants or PRODUCtS/REACTANTS

last one ya on ya own