Question

A solution is prepared to be 0.1475 M in Sr(OH)2. How do you find the [H3O+], [OH‐], pH, and pOH?

Answer 1

`pH = 13.029" and " pOH = 0.9710`

Explanation:

Strontium Hydroxide ionizes in water to produce the strontium ion and the hydroxide ion. Note that `Sr(OH)_2` is slightly soluble in water.

`" "Sr(OH)_2 \ \ rightleftharpoons \ \ \ Sr^(2+) + \ \ \ 2OH^-`

`I " " 0.1475 \ M " " - " " -`

`C" " -x " "+x " "+ 2x`

`E" "0.1475-x " " x " " 2x`

`K_b=( [ OH^-]^2 [ Sr^(2+)])/([Sr(OH)_2]`

`K_b= ( (2x)^2*(x))/(0.1475-x)= 6.5xx10^-3`

Solve for `x`

`x = 0.05346 \ M`

`[ OH^-] = 2x= 2xx 0.05346 = 0.1069 M`

`-------------------`

In any aqueous solution, both ions `H_3O^+`and `OH^-` are present and they must satisfy the following condition:

`[H_3O^+] [ OH^-] = K_W`

`[H_3O^+] [ OH^-] = 1.0xx10^(-14)`

`-------------------`

`[OH^-] = 0.1069 \ M` ( determined)

`[H_3O^+] = (1.0xx10^-14)/([ OH^-] )`

`[H_3O^+] = (1.0xx10^-14)/(0.1069 )`

`[H_3O^+] = 9.355^-14 \ M`

`-------------------`

`pH =- log [H_3O^+]`

`pH =- log [\ 3.955^-14 ]`

`pH = 13.0290`

`pOH =- log [\ OH^-]`

`pOH =- log [\ 0.1069\ ]`

`pOH = 0.9710`