Question

Question #ba54f

Answer 1

`(["B"])/(["BH"^(+)]) = 0.0281`

Explanation:

There are two ways in which you can approach this problem, one using the base dissociation constant and the solution's `pOH`, and the other one using the acid dissociation constant and the solution's pH.

I'll show you how to solve it using `K_b` and `["OH"^(-)]`, and you try the other approach as practice. So, you know that you're dealing with a weak base that has the base dissociation constant, `K_b`, equal to `8.91 * 10^(-6)`.

The equilibrium dissociation of the weak base, which I'll call `"B"` for simplicity, looks like this

`"B"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons "BH"_text((aq])^(+) + "OH"_text((aq])^(-)`

By definition, the base dissciation constant is equal to

`K_b = (["BH"^(+)] * ["OH"^(-)])/([B])`

Use the blood's pH to determine what the `pOH` is

`pH_"sol" + pOH = 14 implies pOH = 14 - pH_"sol"`

`pOH = 14 - 7.4 = 6.6`

The concentration of the hydroxide ions present in solution will be

`["OH"^(-)] = 10^(-pOH)`

`["OH"^(-)] = 10^(-6.6) = 2.5 * 10^(-7)"M"`

Rearrange the equation for `K_b` to get

`K_b * ["B"] = ["BH"^(+)] * ["OH"^(-)]`

`(["B"])/(["BH"^(+)]) = (["OH"^(-)])/K_b = (2.5 * 10^(-7))/(8.91 * 10^(-6)) = color(green)(0.0281)`