Calculate the pH of a solution made by mixing 50 mL of `0.01 Mba(OH)_(2)` with 50 mL water. (Assume complete ionisation)

Explanation

Milli-mole of `Ba(OH)_(2)=50xx0.01=0.5`. The solution is diluted with `50 mL H_(2)O` and thus volume becomes 100 mL.
Thus `[Ba(OH)_(2)]=(0.5)/(100)=0.005M`
`{:(,Ba(OH)_(2)rarr,Ba^(2+)+,20H^(-)),("Initial conc", 0.005,0,0),("Final conc".,0,0.005,2xx0.005=0.01):}`
`[OH^(-)]=0.01=1xx10^(-2)`
`:. pOH=2`
`:. pH=14-2=12`