Question

Question #5f837

Answer 1

`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))`