Question

what is the ph of 0.1 m solution of Ba(OH)2?​

Answer 1

Given :

0.1 M Ba(OH)₂

To find : pH

Formula used :

• pOH = - log[OH⁻]
• pH + pOH = 14

Solution :

Ba(OH)₂ is a strong base , hence it will ionise completely.

Ba(OH)₂ ➝ Ba²⁺ + 2 OH⁻

0.1 M ...............0...............0

0...................(0.1 M).......(2×0.1 M).

➝ This means 0.1M Ba(OH)₂ will give 0.2M OH⁻

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$\sf \implies \: pOH \: = \: - \: log [0.2] \\ \\ \: \sf \implies \: pOH = - \: (log \dfrac{2}{10} ) \\ \\ \sf \implies \: pOH \: = \: - \: ( \: log(2) \: - \: log(10) \: \: ) \\ \\ \sf \implies \: pOH \: = \: - (0.3010 \: - \: 1 \: ) \\ \\ \sf \implies \: pOH = \: - ( - 0.699) \\ \\ \sf \implies \: pOH \: = \: 0.699$

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Now , pH + pOH = 14

This gives,

➝ pH = 14 - pOH

➝ pH = 14 - 0.699

➝ pH = 13.301

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ANSWER :

pH = 13.301