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.