Question

Assuming `"Ba"("OH")_2` dissociates completely, what is the `"pH"` of `"0.015 M"` `"Ba"("OH")_2` to three sig figs?

Answer 1

12.2

Explanation:

`pOH=−log(OH^−)`

Arrhenius acid-base reaction

`Ba(OH)_2 -> Ba^"+2"+2OH^−`

`(OH^−) = 2(0.0075M) = 0.015M(OH^-)`

pOH= −log(0.015) = −(−1.82) = 1.82

pH + pOH= 14

14−1.82 = 12.2

Answer 2

`Ba(OH)_2 rightleftharpoons Ba^(+2) + 2OH^-`

Given 0.00750M `Ba(OH)_2`
=> `[OH^-] = 2(0.00750M) = 0.015M( OH^-)`

=> `pOH=-log[OH^-] = -log(0.015) = -(-1.82) = 1.82`

=> `pH = 14 - pOH = 14 - 1.82 = 12.2`