Question

What is the h oh ph and poh of a 0.005m solution of calcium hydroxide?

Answer 1

Answer: $[OH^{-}]= 0.01M or 1.0\times 10^{-2}M$

$[H^{+}]= 1.0\times 10^{-12}M$

pH = 12

pOH = 2

Explanation: Calcium hydroxide ($Ca(OH)_{2}$) is a strong base that dissociates completely.

Dissociation equation of Calcium hydroxide is :

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

1. Concentration of [OH-]

1 mol $Ca(OH)_{2}$ produces 2 mol OH- ions.

The given solution is 0.005M $Ca(OH)_{2}$ , then  concentration of OH- would be twice the concentration of $Ca(OH)_{2}$

$[OH^{-}] = 0.005\times 2$ = 0.01M or $1.0\times 10^{-2}M$

2.Concentration of [H+]

Concentration of [H+] can be calculated by the formula: $[H^{+}] = \frac{Kw}{[OH^{-}]}$

kw = ionic product of water and its values is $(1\times 10^{-14})$

[OH-] = 0.01 M or $1.0\times 10^{-2}M$

$[H^{+}] = \frac{(1\times 10^{-14})}{[OH^{-}]}$

$[H^{+}] = \frac{(1\times 10^{-14})}{[0.01]}$

$[H^{+}] = 1.0\times 10^{-12}M$

3. pH value

pH is calculated by the formula : $pH = -log[H^{+}]$

$pH = -log[1.0\times 10^{-12}]$

pH = 12

4. pOH value

pOH is calculated by the formula : pOH = 14 - pH

pOH = 14 - 12

pOH = 2

pOH can also be calculated by using a different formula which is :

$pOH = -log(OH^{-})$

$pOH = -log(0.01)$

pOH = 2.