Question #71432

The pH of that solution will be 4.82.

To solve this problem you're going to need the acid dissociation constant, #K_a#, for the ammonium ion, #NH_4^(+)#, which is listed as being equal to #5.6 * 10^(-10)#.

The ammonium chloride will dissociate in aqueous solution to give ammonium, #NH_4^(+)#, and chloride, #Cl^(-)#, ions.

#NH_4Cl_((s)) -> NH_(4(aq))^(+) + Cl_((aq))^(-)#

The ammonium ion will act as a weak acid, reacting with water to produce ammonia and hydronium ions. Use an ICE table to solve for the concentration of the hydroxide ions in solution.

#" "NH_(4(aq))^(+) + H_2O_((l)) rightleftharpoons NH_(3(aq)) + H_3O_((aq))^(+)#
I.......0.42......................................0.................0
C.......(-x).......................................(+x)............(+x)
E....0.42-x......................................x.................x

By definition, the acid dissociation constant will be equal to

#K_a = ([NH_3] * [H_3O^(+)])/([NH_4^(+)]) = (x * x)/(0.42 - x) = x^2/(0.42 - x)#

Because #K_a# is so small, you can approximate (0.42 - x) with 0.42. This will result in

#x^2/0.42 = 5.6 * 10^(-10) => x = 1.53 * 10^(-5)#

This will be equal to the concentration of the hydronium ions

#[H_3O^(+)] = 1.53 * 10^(-5)"M"#

As a result, the pH of the solution will be

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

#pH_"sol" = -log(1.53 * 10^(-15)) = color(green)(4.82)#