Question #da59d

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Question #da59d

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Answer 1 · 2015-01-07T10:41:54

The answer is $1.8 \cdot 10^{- 7}$ $1.8 * 10^(-7)$ $M$ $"M"$ .

The general reaction looks like this

$M g_{(a q)}^{2 +} + 2 N a O H_{(a q)} \to M {g (O H)}_{2 (s)} + 2 N a^{+} (a q)$ $Mg_((aq))^(2+) + 2NaOH_((aq)) -> Mg(OH)_(2(s)) + 2Na^(+)(aq)$

The reaction that is of interest is actually this

$M g (O H) 2 (s) ⇌ M g_{(a q)}^{2 +} + 2 O H_{(a q)}^{-}$ $Mg(OH)2(s) rightleftharpoons Mg_((aq))^(2+) + 2OH_((aq))^(-)$

In order to determine the maximum concentration of $M g^{2 +}$ $Mg^(2+)$ ions permissible in the $N a O H$ $NaOH$ solution before a precipitate will be formed, you'd need the value of the solubility product constant, $K_{s p}$ $K_(sp)$ .

The expression for $K_{s p}$ $K_(sp)$ for this reaction is

$K_{s p} = [M g^{2 +}] \cdot {[O H^{-}]}^{2}$ $K_(sp) = [Mg^(2+)] * [OH^(-)]^2$

A precipitate will form when the solution is saturated with ions; if the concentrations of the ions are low enough, a precipitate will not form. The threshold for the formation of a precipitate implies that the above equation be satisfied.

For a saturated solution, on the verge of forming a precipitate, the maximum value for the product of the concentrations of the two ions is equal to $K_{s p}$ $K_(sp)$ .

From this point on, any increase in the concentrations of the ions will result in the formation of the precipitate.

Since your reaction takes place in 1 L, 0.01 M $N a O H$ $NaOH$ solution, and knowing that $N a O H$ $NaOH$ dissociates completely into $N a^{+}$ $Na^(+)$ and $O H^{-}$ $OH^(-)$ ions, the concentration of hydroxide ions will be

$[O H^{-}] = [N a O H] = 0.01$ $[OH^(-)] = [NaOH] = 0.01$ $M$ $"M"$

This means that the maximum concentration of $M g^{2 +}$ $Mg^(2+)$ ions the solution can hold before a precipitate is formed is

$[M g^{2 +}] = \frac{K_{s p}}{{[O H^{-}]}^{2}} = \frac{K_{s p}}{{(0.01)}^{2}} = \frac{K_{s p}}{10^{- 4}}$ $[Mg^(2+)] = K_(sp)/([OH^(-)]^2) = K_(sp)/(0.01)^2 = K_(sp)/(10^(-4)$

As a numerical value, I've found $K_{s p}$ $K_(sp)$ to be equal to $1.8 \cdot 10^{- 11}$ $1.8*10^(-11)$ (you can use any value given to you, of course), which means that

$[M g^{2 +}] = \frac{1.8 \cdot 10^{- 11}}{10^{- 4}} = 1.8 \cdot 10^{- 7}$ $[Mg^(2+)] = (1.8 * 10^(-11))/10^(-4) = 1.8 * 10^(-7)$ $M$ $"M"$

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