Answer to Question #146006 in Chemistry for Loz

Question

​A solution is made by mixing 500 mL of 0.1 2M NaOH solution with 500 mL of 0.1M. Mg(NO3)2
Ksp for Mg(OH)2. =1.8 x10​

What will Mg^+2 and OH^- be after mixing?

Write down the value of the ionic product quotient

Does a precipitate form?

Answer 1

2NaOH+Mg(NO₃)₂=Mg(OH)₂+2NaNO₃

K_sp for Mg(OH)₂ =1.8 x10^-11​ mol/l

C_M=n/V

n (NaOH) = 0.12 x 0.5 = 0.06 mol

n (Mg(NO₃)₂) =0.1 x 0.5 = 0.05 mol

NaOH is the limiting reagent: it will react fully, 0.2 mol of Mg(NO₃)₂ will be left.

OH^/ iond will come in the resulting solution will come from the resulting Mg(OH)₂ and is eqal to the initial NaOH - 0.06 mol/l.

The concentration of Mg²⁺ will be 0.03 mol/l.

Ionic product equivalent of Mg(OH)₂ is 0.03 x 0.06² = 0.0001

So the precipitate will be formed.