What is the solubility of m(oh)2 in a 0.202 m solution of m(no3)2? ksp 4.45*10^-12?

M(NO₃)₂ fully separates into M²⁺ and NO₃ ²⁻ and M(OH)₂ partially separates as M²⁺ and 2OH⁻

M(NO₃)₂     →        M²⁺      +          2NO₃²⁻

0.202 M                 0.202 M

    M(OH)₂(s)        ↔        M²⁺ (aq) +          2OH⁻(aq)

I                                         -                            -

C   -X                                +X                        +2X

E                                        X                          2X

Ksp                  =          [M²⁺ (aq)] [OH⁻(aq)]²

4.45 * 10∧-12        =          (0.202 + X ) (2X)²

Since X is very small, (0.202 + X ) = 0.202

4.45 * 10-12        =          0.202 * 4X²

            X          =          2.347 × 10∧-6 M

Hence the solubility of M(OH)2 is 2.347 × 10∧-6 M

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.