Question #17cc3

1 Answer
Apr 13, 2015

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")