What is the cobalt (II) ion concentration in a solution prepared by mixing 379 mL of 0.372 M cobalt (II) nitrate with 455 mL of 0.209 M sodium...

The
balanced chemical equation for the reaction between cobalt (II) nitrate and sodium hydroxide can
be written as:

`Co(NO_3)_2 (aq) + 2NaOH (aq) -> Co(OH)_2 (s) + 2NaNO_3
(aq)`

Here, 1 mole of cobalt nitrate reacts with 2 moles of sodium hydroxide
to form 1 mole of cobalt hydroxide and 2 moles of sodium nitrate. 

We are
given that 379 ml of 0.372 M cobalt nitrate is mixed. This means that 0.141 moles (= 0.379 lt x
0.372 moles/lt) of cobalt nitrate has been mixed.

Also, 455 ml of 0.209 M
sodium hydroxide has been used, that is, 0.095 moles (= 0.455 lt x 0.209 moles/lt) has been
mixed.

Using stoichiometry, sodium hydroxide is the limiting reactant and we
will use only 0.095/2 moles = 0.048 moles of cobalt nitrate and will generate only 0.048 moles
of cobalt hydroxide. 

This means that 0.093 moles (= 0.141 - 0.048 moles) of
cobalt nitrate are unused and are in aqueous form. 

The solubility product of
cobalt nitrate is given as 1.3 x 10^-15.

That is, `[Co^(2+)][OH^-]^2 = 1.3 xx
10^(-15) = [Co^(2+)]^3`

solving this, we get, `[Co^(2+)] = 1.091 xx 10^(-5)
M`

Thus, the total concentration of cobalt (II) ions in solution = 0.093
moles + 1.091 x 10^-5 moles = 0.093 moles.

Hope
this helps.