Question #9f499
1 Answer
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,
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
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) = ["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
#color(darkgreen)(ul(color(black)(K_"sp" = 1.5 * 10^(-12))))#