Question

calculate pH of 0.01M of Ba(OH₂) solution​

Answer 1

H+ ions present in HCl solution:  20.0/1000 * 0.050 = 0.001 moles

There are two OH- ions in a molecule of Ba (OH)2.

OH- ions released in the Barium Hydroxide solution =

       = 2 * 30.0/1000 * 0.10 = 0.006 moles

So 0.001 moles of H+ and OH- combine to form water.

So 0.005 moles of OH- remain. So the molarity of [ OH- ]

           = 0.005 moles / (20+30) ml   =0.100 M

Answer 2

Answer:

The pH of 0.01M of Ba(OH)₂ solution is equal to 12.302.

Explanation:

Given: concentration of Ba(OH)₂ solution $=0.01M$

concentration of OH⁻ ions of Ba(OH)₂ solution $=2*0.01=0.02moles$

pH scale is used to find the acidic and basic strength of solution.

We know that  $pH+pOH=14$

Therefore first find pOH:

We have [OH]⁻ = 0.02 moles

Therefore, $pOH=-log[OH^{-}]$

⇒    $pOH=-log[0.02]$

⇒     $pOH=-(-1.698)$

∴    $pOH=1.698$

Now, $pH=14-pOH$

        $pH=14-1.698$

⇒      $pH=12.302$

Therefore the pH of 0.01M of Ba(OH)₂ solution is 12.302.