Question

⦁ 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.

Answer 1

`pH=12.3\\\\pOH=1.7\\`

`[H^+]=5x10^(-13)M`

`[OH^-]=0.02M`

Step-by-step 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:

`Mg(OH)_2(s)\rightleftharpoons Mg^(2+)(aq)+2OH^-(aq)`

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

`(1)/(3) =(x)/([Mg(OH)_2])`

Thus, x for this problem is:

`x=([Mg(OH)_2])/(3)=(0.03M)/(3)\\\\x= 0.01M`

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

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

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

`[H^+]=(1x10^(-14))/(0.02)=5x10^(-13)M`

`pH=-log(5x10^(-13))=12.3\\\\pOH=14-pH=14-12.3=1.7`

Best regards!