Question

Calculate the ph of 0.05M ba(OH)2 aqueous solution

Answer 1

$ASSALAMUALAIKUM$

[OH-] = N = 2×0.05 = 0.1

pOH = -log[OH-] = -log(0.1) = 1

PH = 14-pOH = 14-1 = 13

$INSHAALLAH it will help you!$

Answer 2

The pH of 0.05M $Ba(OH)_2$ aqueous solution is 13.

Given:

Molarity of the solution = 0.05M

To find:

pH of the solution = ?

Formula to be used:

pOH = -log($OH^-$)

pH = 14 - pOH

Calculation:

In $Ba(OH)_2$, 2 hydroxide ions are present.

∴ 0.05 M = [$Ba(OH)_2$]

[$OH^-$] = 2 ×0.05 = 0.1M

∴pOH =  -log($OH^-$) = -log(0.1) = 1

Substituting the obtained value in the following formula it gives

pH = 14 - pOH

pH = 14 - 1

pH = 13

Conclusion:

The pH of 0.05M  $Ba(OH)_2$ aqueous solution is calculated as 13.

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