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* 10^(-2)M`

`[H^(+)]= 1.0* 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* 2` = 0.01M or
`1.0* 10^(-2)M`

2.Concentration of [H+]

Concentration of [H+] can be calculated by the formula:
`[H^(+)] = (Kw)/([OH^(-)])`

kw = ionic product of water and its values is
`(1* 10^(-14))`

[OH-] = 0.01 M or
`1.0* 10^(-2)M`

`[H^(+)] = ((1* 10^(-14)))/([OH^(-)])`

`[H^(+)] = ((1* 10^(-14)))/([0.01])`

`[H^(+)] = 1.0* 10^(-12)M`

3. pH value

pH is calculated by the formula :
`pH = -log[H^(+)]`

`pH = -log[1.0* 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.