Calculate the #pH# at which #Mg(OH)_2# begins to precipitate from a solution containing #0.1# #M# #Mg^(2+)# ions? #K_(sp)# for #Mg(OH)_2# = #1.0# x #10^-11#.

Question

Wyatt Atkins · Accepted Answer

#"pH" = 9#

The idea here is that you need to use magnesium hydroxide's solubility product constant to determine what concentration of hydroxide anions, #"OH"^(-)#, would cause the solid to precipitate out of solution.

As you know, the dissociation equilibrium for magnesium hydroxide looks like this

#"Mg"("OH")_text(2(s]) rightleftharpoons "Mg"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-)#

The solubility product constant, #K_(sp)#, will be equal to

#K_(sp) = ["Mg"^(2+)] * ["OH"^(-)]^color(red)(2)#

Rearrange to find the concentration of the hydroxide anions

#["OH"^(-)] = sqrt(K_(sp)/(["Mg"^(2+)]))#

Plug in your values to get

#["OH"^(-)] = sqrt((1.0 * 10^(-11))/0.1) = sqrt(1.0 * 10^(-10)) = 10^(-5)"M"#

As you know, you can use the concentration of hydroxide anions to find the solution's pOH

#color(blue)("pOH" = - log(["OH"]^(-)))#

#"pOH" = - log(10^(-5)) = 5#

Finally, use the relationship that exists between pOH and pH at room temperature

#color(blue)("pOH " + " pH" = 14)#

to find the pH of the solution

#"pH" = 14 - 5 = color(green)(9)#

So, for pH values that are below #9#, the solution will be unsaturated. Once the pH of the solution becomes equal to #9#, the solution becomes saturated and the magnesium hydroxide starts to precipitate.

Jose Ayala · Answer

undefined To find the pH at which $ \text{Mg(OH)}_2 $ begins to precipitate, we first need to write down the dissolution equation and its solubility product constant ($ K_{sp} $) expression.

$$ \text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^{-}(aq) $$

The $ K_{sp} $ for $ \text{Mg(OH)}_2 $ is given as $ 1.0 \times 10^{-11} $, and this is equal to the concentrations of the ions at equilibrium:

$$ K_{sp} = [ \text{Mg}^{2+} ][ \text{OH}^{-} ]^2 = 1.0 \times 10^{-11} $$

Given that the concentration of $ \text{Mg}^{2+} $ ions is $ 0.1 \, \text{M} $, we can substitute this value into the equation to solve for $ [ \text{OH}^{-} ] $:

$$ 1.0 \times 10^{-11} = (0.1)([ \text{OH}^{-} ])^2 $$

$$ [ \text{OH}^{-} ]^2 = \frac{1.0 \times 10^{-11}}{0.1} $$

$$ [ \text{OH}^{-} ]^2 = 1.0 \times 10^{-10} $$

$$ [ \text{OH}^{-} ] = \sqrt{1.0 \times 10^{-10}} $$

Let's calculate the $ [ \text{OH}^{-} ] $.The concentration of $ \text{OH}^{-} $ ions is $ 1.0 \times 10^{-5} \, \text{M} $.

To find the pH, we first calculate the pOH using the $ [ \text{OH}^{-} ] $:

$$ \text{pOH} = -\log([ \text{OH}^{-} ]) $$

Then, we use the relationship between pH and pOH:

$$ \text{pH} + \text{pOH} = 14 $$

Let's calculate the pH.The pH at which $ \text{Mg(OH)}_2 $ begins to precipitate from a solution containing $ 0.1 \, \text{M} $ $ \text{Mg}^{2+} $ ions is $ 9.0 $.