The concentration of Mg2+ in seawater is 0.052 M. At what pH will 76% of the Mg2+ be precipitated as the hydroxide salt? [Ksp for Mg(OH)2 = 8.9 ✕ 10-12.]

Answer:

At pH 4.85 , 76% of [tex]Mg^2^+[/tex]will be precipitated as the hydroxide salt.

Explanation:

When 76% of [tex]Mg^{2}^{+}[/tex] has precipitated out , 24% remains out.

Given= Ksp for [tex]Mg(OH_2) = 8.9\times10^-^1^2[/tex]

(Here , ksp is the molar solubility )

Now ,

   [tex][Mg^{2}^{+}] =\frac{24}{100}\times 0.052[/tex]

             = [tex]1.248\times10^-^2[/tex]

   [tex]Mg(OH)_2<=> Mg^2^+ + 2OH^-[/tex]

We , know the formula to find out the ksp ,

Given formula is -

     [tex]Ksp = [Mg^2^+][OH^-][/tex]

Now , putting the given values ,

[tex]8.9\times 10^-^1^2= 1.248\times10^-^2\times[OH]^-[/tex]

    [tex][OH]^- = 7.131\times 10^-^1^0[/tex]

Now , calculating the pH,

[tex][H^+]=\frac{Kw}{[OH]^-}[/tex] ( where Kw is the ionic product of the water)

          [tex]= \frac{10^-^1^4}{7.131\times10^-^1^0}[/tex]

             [tex]1.402\times10^-^5[/tex]

 [tex]pH=-Log[H^+][/tex]

          [tex]-Log[1.402\times10^-^5][/tex]

         = 4.85

Therefore , the pH to be precipitated is 4.85

pH measures the acidic or the basic concentration of the substance in a solution. At pH 4.85, the magnesium ion will precipitate as its hydroxide salt.

The potential of hydrogen is calculated as the negative log of the hydrogen ion concentration in the aqueous solution.

The balanced reaction is shown as,

[tex]\rm Mg(OH_{2}) \Leftrightarrow Mg^{2+} + 2OH^{-}[/tex]

If, 76% magnesium precipitates then 24% will remain in. The concentration of the remaining magnesium will be,

[tex]\begin{aligned} \rm [Mg^{2+}] &= \dfrac{24}{100}\times 0.052\\\\&= 1.248 \times 10^{-2}\end{aligned}[/tex]

The molar solubility is calculated as:

[tex]\begin{aligned} \rm Ksp &= \rm [Mg^{2+}][OH^{-}]\\\\8.9 \times 10^{-12} &= 1.248 \times 10^{-2} \times \rm [OH^{-}]\\\\&= 7.13 \times 10^{-10}\end{aligned}[/tex]

The concentration of the hydrogen ions is calculated as:

[tex]\begin{aligned} \rm [H^{+}] &= \rm \dfrac{Kw}{[OH^{-}]}\\\\&= \dfrac{10^{-14}}{ 7.13 \times 10^{-10}}\\\\&= 1.402 \times 10^{-5}\end{aligned}[/tex]

Further, pH is calculated as:

[tex]\begin{aligned} \rm pH &=\rm -log [H^{+}]\\\\&= -\rm log [1.402 \times 10^{-5}]\\\\&= 4.85\end{aligned}[/tex]

Therefore, at 4.85 pH the magnesium ion will precipitate out as hydroxide salt.

Learn more about pH and precipitation here:

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