Question #5f837

1 Answer
Feb 11, 2016

Answer:

#1.0 * 10^(-8)#

Explanation:

Start by writing the balanced chemical equation for this hydrolysis reaction

#"B"_text((aq])^(+) + "H"_2"O"_text((l]) rightleftharpoons "BOH"_text((aq]) + "H"_text((aq])^(+)#

The **equilibrium constant, #K_(eq)#, for this reaction is defined as

#K_(eq) = (["BOH"] * ["H"^(+)])/(["B"^(+)] * ["H"_2"O"])#

Now, because water's concentration is assumed to be constant, you can rewrite this equation as

#overbrace(K_(eq) * ["H"_2"O"])^(color(purple)(=K_h)) = (["BOH"] * ["H"^(+)])/(["B"^(+)])#

The left-hand side of the equation is equal to the hydrolysis constant ,#K_h#, so you can say that

#K_h = color(blue)((["BOH"])/(["B"^(+)])) * ["H"^(+)]" " " "color(red)("(*)")#

You are told that the base dissociation constant, #K_b#, for this reaction is equal to #1.0 * 10^(-6)#. The balanced chemical equation for the partial dissociation of the base looks like this

#"BOH"_text((aq]) rightleftharpoons "B"_text((aq])^(+) + "OH"_text((aq])^(-)#

By definition, #K_b# will be equal to

#K_b = (["B"^(+)] * ["OH"^(-)])/(["BOH"])#

Rearrange this to get

#K_b/(["OH"^(-)]) = (["B"^(+)])/(["BOH"])#

#color(blue)((["BOH"])/(["B"^(+)])) = (["OH"^(-)])/K_b#

Plug this into equation #color(red)("(*)")# to get

#K_h = (["OH"^(-)])/K_b * ["H"^(+)]#

#K_h = (["OH"^(-)] * ["H"^(+)])/K_b#

You also know that the ion product constant for water, #K_W#, equal to #10^(-14)# at room temperature, is defined as

#K_W = ["H"^(+)] * ["OH"^(-)]#

This means that you have

#K_h = K_W/K_b#

Plug in your value to get

#K_h = 10^(-14)/(1.0 * 10^(-6)) = color(green)(1.0 * 10^(-8))#