20g of sample containing `Ba(OH)_(2)` is dissolved in 10 ml of 0.5 M HCl solution. The excess of HCl was then titrated against 0.2 M NaOH. The volume of NaOH used in the titration was 10 ml. Calculate the percentage of `Ba(OH)_(2)` in the sample. (Mol. wt. of `Ba(OH_(2))=171)`

Explanation

Calculation of volume of HCl used in titration between NaOH and HCl.
`V_(NaOH) =10 ml, M_(NaOH) = 0.2 M, M_(HC) = 0.5 M, V_(HCl)`=?
`M_(NaOH) xx V_(NaOH) = M_(HCl) xx V_(HCl), V_(HCl) = (10 xx 0.2)/0.5 = 4 ml`
This is the volume of HCl left unused when excess of HCl is added to `Ba(OH)_(2)` solution.
Total volume of HCl added = 10 ml
Volume of HCl used to react with `Ba(OH)_(2) = 10-4 = 6` ml
`2HCl + Ba(OH)_(2) to BaCl_(2) + 2H_(2)O`
M = No. of moles x `1000/("volume of solution") rArr 0.5 = "moles" xx 1000/6`.
`(0.5 xx 6)/1000` = moles of HCl Moles of HCl used = 0.003 moles.
Observing the molar ratio of HCl and `Ba(OH)_(2)`.
Moles of `Ba(OH)_(2)` reacted `=1/2` x moles of HCl reacted `=1/2 xx 0.003 = 0.0015` moles.
Weight of `Ba(OH)_(2)` reacted = no. of moles x mol. wt. =` 0.0015 xx 171 = 0.2565` g
Percentage of `Ba(OH)_(2)` in the sample `=(wt. of Ba(OH)_(2) "reacted")/("Total weight of sample") xx 100` = 1.28 %