Question

Find the pOH and pH of 0.2N Ba(OH)2?​

Answer 1

Answer:

molarity = Normality/n-factor

Normality = molarity×n-factor

0.2 = molarity× 2

molarity = 0.2/2 = 0.1 M

Ba(OH)2 = Ba^2+ + 2 OH^-

0.1M. 2×0.1M

[OH-] = 0.2M

pOH = -log[OH-]

= - log (2×10^-1)

=. 1-log2

=. 1- 0.3010

= 0.699

now PH + POH =14

PH = 14 - 0.699

PH= 13.301

Answer 2

Answer:

pOH = 1

pH = 13

Given:

Normality of Ba(OH)₂, N = 0.2 N

Find:

pOH and oH of the solution.

Solution:

Since Ba(OH)₂ is a very strong base, it dissociate completely in its aqueous solution.

Ba(OH)₂ --------------> Ba²⁺ + 2OH⁻

1 mole of Ba(OH)₂ gives 2 OH⁻ ions.

∴ Molarity of OH⁻ ions, M = Normality/n-factor

                                           = 0.2/2 = 0.1 M

We know that,

pOH = log₁₀ [OH⁻]

where [OH⁻] = Concentration of OH⁻ ions = Molarity of OH⁻ ions

∴ pOH = log₁₀ (0.1)

pOH = log₁₀ (1/10) = log₁₀ 1 - log₁₀ 10

pOH = 0 - 1  = -1 = 1            [ Ignoring negative sign]

Hence, pOH of 0.2 N Ba(OH)₂ is 1.

As we know,

pOH + pH = 14

pH = 14 - pOH

pH = 14 - 1

pH = 13

Hence, pH of 0.2 N Ba(OH)₂ is 13.

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