What will be the pH of the solution obtained when 50cc of 0.1(M) `Ca(OH)_(2)` solution is diluted to 500cc by ading water into it?

Explanation

In 50cc 0.1(M) `Ca(OH)_(2)` solution, number of millimoles of `Ca(OH)_(2)=0.1xx50=5`. [Number of millimoles=molarity of the solution`x`volume of the solution (mL)].
this amount of `Ca(OH)_(2)` is also present in 500cc solution. So, the concentration of 500 cc solution `=(5)/(500)=10^(-2)(M)`. in aqueous solution, `Ca(OH)_(2)` undergoes complete ionisation and each `Ca(OH)_(2)` molecule gives two `OH^(-)` ions on ionisation.
Therefore, the concentration of `OH^(-)` ions in 500cc solution`=2xx10^(-2)(M)`.
`therefore pOH` of the solution`=-log_(10)(2xx10^(-2))=1.69 and pH=14-1.69=12.31`