Question #9f499

1 Answer
Nov 13, 2016

Answer:

#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))))#