Question

A solution is made by mixing 500.0ml of 0.12M NaOH soulution with 500mL of 0.1M Mg(NO3)2

Ksp for Mg =1.8 x10 x^-11
Does a precipitate form?

Answer 1

The chemical equation for the reaction is;

$Mg(NO_3)_{2_(aq)}+2NaOH_{(aq)}\to Mg(OH)_{2(s)}+2NaNO_{3(aq)}$

The ionic equation for reaction reaction is;

$Mg^{2+}+2NO_3^{-}+2Na^++2OH^{-}\to2Na^++2NO_3^-+Mg(OH)_2$

Net ionic equation is;

$Mg(OH)_2 \leftrightharpoons Mg^{2+}+2OH^-$

$K_{sp}$ $=[Mg^{2+}][OH^-]^2$ $=1.8\times 10^{-11}$

Molar mass of$Mg(OH)_2=58.319$

$Mg(OH)_2=$ $1.8\times 10^{-11}gMg(OH)_2\over1L$ $\times$ $2mol Mg(OH)_2\over58.319gMg(OH)_2$ $=$ $6.173\times10^{-13}$

There is formation of a presipitate.