⦁ Find the concentration of H+, OH-, PH and POH of 0.03 M of magnesium hydroxide which ionizes to the extent of only 1 /3 in aqueous solution.

Answers

Answer 1

Answer:

[tex]pH=12.3\\\\pOH=1.7\\[/tex]

[tex][H^+]=5x10^{-13}M[/tex]

[tex][OH^-]=0.02M[/tex]

Explanation:

Hello there!

In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:

[tex]Mg(OH)_2(s)\rightleftharpoons Mg^{2+}(aq)+2OH^-(aq)[/tex]

Thus, since the ionization occurs at an extent of 1/3, we can set  up the following relationship:

[tex]\frac{1}{3} =\frac{x}{[Mg(OH)_2]}[/tex]

Thus, x for this problem is:

[tex]x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x= 0.01M[/tex]

Now, according to an ICE table, we have that:

[tex][OH^-]=2x=2*0.01M=0.02M[/tex]

Therefore, we can calculate the H^+, pH and pOH now:

[tex][H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M[/tex]

[tex]pH=-log(5x10^{-13})=12.3\\\\pOH=14-pH=14-12.3=1.7[/tex]

Best regards!


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