Question

What is the poh of a 0.02m ca(oh)2.

Answer 1

pOH=40

pOH= -log10 [OH^-]

=0.04 mol

= -log10(0.04)=-(-1.40)

=1.4

Answer 2

Answer:

The pOH of 0.02M solution of Ca(OH)₂ is equal to 1.397.

Explanation:

Given: the concentration of Ca(OH)₂ $=0.02M$

Calcium hydroxide is a strong base but it is not completely soluble. But because of strong base, it will ionize completely.

                        $Ca(OH)_{2}$   →   $Ca^{2+}$     +     $2OH^{-}$

Initial conc.       0.02M           ------             -------

Final conc.        --------           0.02M          0.04M

So, the concentration of $[OH]^{-}$ions in the solution $=0.04M$

Now, pOH = -log [OH⁻]

⇒    $pOH= -log[0.04]$

⇒   $pOH=1.397$

Therefore, the pOH of 0.02M calcium hydroxide solution is 1.397.