20g of a sample of `Ba(OH)_(2)` is dissolved in 10ml of `0.5N HCI` solution. The excess of HCI was titrated with `0.2N NaOH`. The volume of NaOH used was 10 c c. Calculate the percentage of `Ba(OH)_(2)` in the sample.

Explanation

Milli eq. of HCI initially `= 10 xx 0.5 =5`
Milli eq. of NaOH consumed = Milli eq. of HCI is excess `= 10 xx 0.2 =2`
`:.` Milli eq. of HCI consumed = Milli eq. of `Ba(OH)_(2) = 5 - 2 = 3`
`:.` eq. of `Ba(OH)_(2) = 3//1000 = 3 xx 10^(-3)`
Mass of `Ba(OH)_(2) = 3 xx 10^(-3) (171//2) = 0.2565g`
`%Ba(OH)_(2) = (0.2565//2) 100 = 1.28%`