Question

the ksp of magnesium fluoride is 3.7x10^-8.When 400.0 ml of 0.600 M NaF is mixed with 600. ml of 0.217 M Mg(NO3)2, a precipitate forms.
(a) what mass of megangsim fluoride(M MASS=62.30g/mol) is produced ?
(b) what will be the equilibrium concentrations go Mg^2+ and F^- in the resulting solution?

Answer 1

Answer on Question #56432 – Chemistry - Other

Question:

The $k_{\mathrm{sp}}$ of magnesium fluoride is $3.7 \times 10^{-8}$. When $400.0 \, \mathrm{ml}$ of $0.600 \, \mathrm{M} \, \mathrm{NaF}$ is mixed with $600 \, \mathrm{ml}$ of $0.217 \, \mathrm{M} \, \mathrm{Mg(NO_3)_2}$, a precipitate forms.

(a) what mass of megangsim fluoride $(M = 62.30 \, \mathrm{g/mol})$ is produced?

(b) what will be the equilibrium concentrations go $\mathrm{Mg}^{2+}$ and $\mathrm{F}^{-}$ in the resulting solution?

Answer:

a) $\mathrm{Mg(NO_3)_2 + 2NaF = 2NaNO_3 + MgF_2}$

According to the reaction, $v(MgF_{2}) = v(Mg(NO_{3})_{2})$

$$\begin{array}{l} v = \frac{m}{M} = C_{M} V \quad m(NaF) = C_{M}(Mg(NO_{3})_{2}) \cdot V(Mg(NO_{3})_{2}) \cdot M(MgF_{2}) \\ \quad m(NaF) = 0.600 \cdot 0.217 \cdot 62.30 = 8.11 \, g \end{array}$$

b) $K_{\mathrm{sp}} = [\mathrm{Mg}^{+}] [\mathrm{F}^{-}]^{2}$

$$\begin{array}{l} 3.7 \times 10^{-8} = [\mathrm{Mg}^{+}] [\mathrm{F}^{-}]^{2} \\ [\mathrm{Mg}^{+}] = (3.7 \times 10^{-8})^{1.5} = 0.003 \, \mathrm{M} \\ [\mathrm{F}^{-}] = 0.003^{2} = 0.00001 \, \mathrm{M} \end{array}$$

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