Question

Question #a3dd1

Answer 1

Your solution will have a pH equal to 4.0.

Since you're dealing with a weak acid, you can use an ICE table on its dissociation equilibrium to determine the equilibrium concentrations of the species involved in the reaction

`" "HA_((aq)) + H_2O_((l)) rightleftharpoons H_3O_((aq))^(+) + A_((aq))^(-)`
I.......1.........................................0.................0
C....(-x)......................................(+x).............(+x)
E.....1-x........................................x.................x

The acid dissociation constant, `K_a`, will be equal to

`K_a = ([H_3O^(+)] * [A^(-)])/([HA]) = (x * x)/(1-x) = x^2/(1-x)`

Since `K_a` has such a small value, you ca approximate (1-x) to be equal to 1, which would mean that

`K_a = x^2/1 = x^2 = 10^(-8) => x = sqrt(10^(-8)) = 10^(-4)`

Since`x` is equal to the equilibrium concentration of the hydronium ion, you'll have

`[H_3O^(+)] = 10^(-4)"M"`

Therefore, the pH of the solution will be

`pH_"sol" = -log([H_3O^(+)])`

`pH_"sol" = -log(10^(-4)) = color(green)(4.0)`