Question

Question #1809d

Answer 1

`K_b = 3.7 * 10^(-6)`

Explanation:

Start by using the known pH of the solution to find the molarity of the hydrogen ions, `"H"^(+)`. This will help you find the acid dissociation constant, `K_a`, for quinine's conjugae acid, `"QH"^(+)`.

So, you know that quinine's conjugate acid, `"QH"^(+)`, will ionize in aqueous solution to give

`"OH"_text((aq])^(+) -> "Q"_text((aq]) + "H"_text((aq])^(+)`

The concentration of the conjugate acid is simply the ratio between the number of moles you have and the volume of the solution

`["QH"^(+)] = "0.23 moles"/"1.0 L" = "0.23 M"`

The concentration of `"H"^(+)` will be

`["H"^(+)] = 10^(-"pH") = 10^(-4.58) = 2.63 * 10^(-5)"M"`

Since the dissociation reactrion produces equal numbers of moles of `"Q"` and `"H"^(+)`, you get that

`["Q"] = ["H"^(+)] = 2.63 * 10^(-5)"M"`

The acid dissociation constant for this equilibrium will thus be

`K_a = (["Q"] * ["H"^(+)])/(["QH"^(+)])`

`K_a = (2.63 * 10^(-5) * 2.63 * 10^(-5))/0.23 = 2.7 * 10^(-9)`

Finally, to get the base dissociation constant, `K_b`, use the fact that

`K_a * K_b = K_W" "`, where

`K_W` - the water self-ionization constant, equal to `10^(-14)`.

This means that `K_b` will be equal to

`K_b = K_W/K_a = 10^(-14)/(2.7 * 10^(-9)) = color(green)(3.7 * 10^(-6))`