Problem: Will a precipitate form when 100.0 mL of 4.0 X 10 -4 M Mg(NO3)2 is added to 100.0 mL of 2.0 X 10 -4 M NaOH?

FREE Expert Solution

For this, we need to compare the reaction quotient (Q) vs. the solubility product constant (Ksp)

Recall that when:

• Q > Ksp: the solution is supersaturated and a precipitate will form. Reactants are favored.

• Q = Ksp: the solution is at equilibrium and no precipitate will form.

• Q < Ksp: the solution is unsaturated and no precipitate will form. Products are favored.

The expected precipitate is Mg(OH)2. The hydroxide ion, OH, has a charge of –1. Mg then has a charge of +2.

The dissociation of Mg(OH)2 in water is as follows:

Mg(OH)2(s)  Mg2+(aq) + 2 OH(aq)

The reaction quotient expression for Mg(OH)is:

$\overline{){\mathbf{Q}}{\mathbf{=}}\frac{\mathbf{products}}{\overline{)\mathbf{reactants}}}{\mathbf{=}}\mathbf{\left[}{\mathbf{Mg}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}{\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}}^{{\mathbf{2}}}}$

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Problem Details

Will a precipitate form when 100.0 mL of 4.0 X 10 -4 M Mg(NO3)2 is added to 100.0 mL of 2.0 X 10 -4 M NaOH?