Question #cda49

The pH of the solution will be equal to 10.1.

Sodium hypochlorite will dissociate in aqueous solution to give sodium cations, #Na^(+)#, and hypochlorite anions, #ClO^(-)#.

The hypochlorite anions will act as a base and react with the water to form hypochlorous acid, #HClO#, and hydroxide ions, #OH^(-)#. You're going to have to determine the base dissociation constant, #K_b#, for the hypochlorite ion.

#K_b = K_w/K_a = 10^(-14)/(4.0 * 10^(-8)) = 2.5 * 10^(-7)#

Now use an ICE table to solve for the concentration of hydroxide ions.

#" "ClO_((aq))^(-) + H_2O_((l)) rightleftharpoons HClO_((aq)) + OH_((aq))^(-)#
I......0.081......................................0.................0
C.......(-x)......................................(+x)..............(+x)
E....0.081-x....................................x.................x

By definition, the base dissociation constant will be

#K_b = ([HClO] * [OH^(-)])/([ClO^(-)]) = (x * x)/(0.081 - x) = x^2/(0.081-x)#

Because the base dissociation constant is so small, you can approximate (0.081 - x) with 0.081. This will get you

#K_b = x^2/0.081 = 2.5 * 10^(-7) => x = 1.42 * 10^(-4)#

This will be equal to the concentration of the hydroxide ions

#[OH^(-)] = 1.42 * 10^(-4)"M"#

The solution's pOH will be

#pOH = -log([OH^(-)])#

#pOH = -log(1.42 * 10^(-4)) = 3.9#

Thus, the solution's pH will be

#pH_"sol" = 14 - pOH = 14 - 3.9 = color(green)(10.1)#