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
The solubility of m(OH)2 in a 2.202 m solution of m(NO³) 2 is [tex]2.347 \times 10^-^6 M[/tex]
Solubility is defined as the amount of solute dissolved in a solvent.
[tex]M(NO_3)_2 = M^2^+ + 2NO_3^2^-[/tex]
0.202 M 0.202 M
[tex]M(OH)_2(s) = M^2^+ (aq) + 2OH^-(aq)[/tex]
Calculating the ICE table
I - - -
C -X +X +2X
E X 2X
[tex]Ksp = [M^2^+(aq)] [OH^-(aq)]^2[/tex]
x is very small, hence (0.202 + X ) = 0.202X
[tex]4.45 \times 10^-^1^2 = (0.202 + X ) (2X)^2[/tex]
[tex]4.45 \times 10^-^1^2 = 0.202\times 4X^2[/tex]
X = [tex]2.347 \times 10^-^6 M[/tex]
Thus, the solubility is [tex]2.347 \times 10^-^6 M[/tex]
Learn more about solubility, here:
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