Calculate the pH of 0.15 M aqueous solution of ammonia?

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    The of the solution is 11.72.
    To solve this problem you need the value of the base dissociation constant of ammonia ##K_b## which is listed as being ##1.8 * 10^(-5)##.
    Since ammonia is aweak base it will increase the concentration of ##OH^(-)## ions in solution so you would expect the solution to have a greater than 7.
    Use the ICE table (more here: of ##OH^(-)## ions – labeled as ##x##
    ## NH_(3(aq)) + H_2O_((l)) rightleftharpoons NH_(4(aq))^(+) + OH_((aq))^(-)##I…….0.15………………………………….0…………….0C……(-x)…………………………………..(+x)………..(+x)E…..0.15-x………………………………..x…………….x
    Use the definition of the base dissociation constant
    ##K_b = ([OH^(-)] * [NH_4^(+)])/([NH_3])##
    ##1.8 * 10^(-5) = ( x * x)/(0.15 – x) = x^2/(0.15-x)##
    Because ##K_b## is so small you can approximate 0.15 – x with 0.15
    ##1.8 * 18^(-5) = x^2/0.15 => x = 0.005196##
    The solution’s pOH will be
    ##pOH = – log([OH^(-)]) = -log(0.005196) = 2.28##
    Therefore the pH of the solution will be
    ##pH_sol = 14 – pOH = 14 – 2.28 = color(green)(11.72)##

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