The answer is A) 2 * 10^(-3)"M"2⋅10−3M.
Because the solubility product constant, K_(sp)Ksp, of manganese (II) hydroxide is so small, the compound is considered insoluble in aqueous solution.
However, very, very little amounts will dissociate according to the balanced equilibrium equation
Mn(OH)_text(2(s]) rightleftharpoons Mn_text((aq])^(2+) + color(red)(2)OH_text((aq])^(-)Mn(OH)2(s]⇌Mn2+(aq]+2OH−(aq]
According to the definition of the solutbility constant product, you have
K_(sp) = [Mn^(2+)] * [OH^(-)]^(color(red)(2))Ksp=[Mn2+]⋅[OH−]2 color(blue)((1))(1)
Every time the concentrations of the Mn^(2+)Mn2+ and OH^(-)OH− ions satisfy the above equation, a precipitate will form.
You can determine the concentration of the hydroxide ions from the solution's pH
pOH = 14 - pH_"sol" = 14 - 9 = 5pOH=14−pHsol=14−9=5
[OH^(-)] = 10^(-pOH) = 10^(-5)[OH−]=10−pOH=10−5
Therefore, according to equation color(blue)((1))(1),
[Mn^(2+)] = K_(sp)/([OH^(-)])^(2) = (2 * 10^(-13))/(10^(-5))^(2) = (2 * 10^(-13))/10^(-10)[Mn2+]=Ksp([OH−])2=2⋅10−13(10−5)2=2⋅10−1310−10
[Mn^(2+)] = color(green)(2 * 10^(-3)"M")[Mn2+]=2⋅10−3M