Calculate the pH of the following solution:
0.3 g of `Ca(OH)_(2)` dissolved in water to give 500 mL of solution.

Explanation

Concentration of `Ca(OH)_(2)`,
`[Ca(OH)_(2)]=(0.3)/(74)xx(1000)/(500)=8.1xx10^(-3)(M)`
[Gram-formulas mass of `Ca(OH)_(2)=74g*mol^(-1)`]
In this solution, `[OH^(-)]=2xx8.1xx10^(-3)(M)=1.61xx10^(-2)(M)`
[Since on ionisation each molecule of `Ca(OH)_(2)` gives two `OH^(-)` ions]
`therefore pOH=-log_(10)[OH^(-)]=-log_(10)(1.61xx10^(-2))=1.8`
and `pH=14-1.8=12.2`