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

in chemistry, a solution’s concentration is how much of a dissolvable substance, known as a solute, is mixed with another substance, called the solvent. the standard formula is c = m/v, where c is the concentration, m is the mass of the solute dissolved, and v is the total volume