Question

Question #9f499

Answer 1

`K_"sp" = 1.5 * 10^(-12)`

Explanation:

Your starting point here is the pH of the solution. More specifically, you need to use the given pH to determine the concentration of hydroxide anions, `"OH"^(-)`, present in the saturated solution.

As you know, an aqueous solution kept at room temperature has

`color(blue)(ul(color(black)("pH " + " pOH" = 14)))`

This means that your solution has

`"pOH" = 14 - 10.16 = 3.84`

Now, the pOH of the solution gives you its concentration of hydroxide anions

`color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))`

To find the concentration of hydroxide anions, rearrange the equation as

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

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

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

Plug in your value to find

`["OH"^(-)] = 10^(-3.84) = 1.445 * 10^(-4)"M"`

Now, your unknown salt only paritally dissolves i nwater, meaning that an equilibrium is established between the dissolves ions and the undissolved solid

`"M"("OH")_ (2(s)) rightleftharpoons "M"_ ((aq))^(2+) + color(red)(2)"OH"_ ((aq))^(-)`

Notice that every mole of salt that dissociates produces `1` mole of `"M"^(2+)` cations and `color(red)(2)` moles of `"OH"^(-)` anions.

This means that the equilibrium concentration of hydroxide anions will be twice as high as that of the metal cations. Therefore, you can say that

`["M"^(2+)] = (["OH"^(-)])/color(red)(2)`

`["M"^(2+)] = (1.445 * 10^(-4)"M")/color(red)(2) = 7.225 * 10^(-5)"M"`

Finally, the solubility product constant, `K_(sp)`, is equal to

`K_(sp) = ["M"^(2+)] * ["OH"^(-)]^color(red)(2)`

Plug in your values to find

`K_(sp) = 7.225 * 10^(-5)"M" * (1.445 * 10^(-4)"M")^color(red)(2)`

`K_(sp) = 1.509 * 10^(-12)"M"^3`

Rounded to two sig figs, the number of decimal places you have for the pH of the solution, and express without added units, the `K_(Sp)` will be

`color(darkgreen)(ul(color(black)(K_"sp" = 1.5 * 10^(-12))))`