Question #5f837
1 Answer
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) = (["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 = color(blue)((["BOH"])/(["B"^(+)])) * ["H"^(+)]" " " "color(red)("(*)")#
You are told that the base dissociation constant,
#"BOH"_text((aq]) rightleftharpoons "B"_text((aq])^(+) + "OH"_text((aq])^(-)#
By definition,
#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
#K_h = (["OH"^(-)])/K_b * ["H"^(+)]#
#K_h = (["OH"^(-)] * ["H"^(+)])/K_b#
You also know that the ion product constant for water,
#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))#