The answer is A) #2 * 10^(-3)"M"#.
Because the solubility product constant, #K_(sp)#, 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])^(-)#
According to the definition of the solutbility constant product, you have
#K_(sp) = [Mn^(2+)] * [OH^(-)]^(color(red)(2))# #color(blue)((1))#
Every time the concentrations of the #Mn^(2+)# and #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 = 5#
#[OH^(-)] = 10^(-pOH) = 10^(-5)#
Therefore, according to equation #color(blue)((1))#,
#[Mn^(2+)] = K_(sp)/([OH^(-)])^(2) = (2 * 10^(-13))/(10^(-5))^(2) = (2 * 10^(-13))/10^(-10)#
#[Mn^(2+)] = color(green)(2 * 10^(-3)"M")#
