20g of sample containing Ba(OH)_(2) is dissolved in 10 ml of 0.5 MHCl 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)

20g of sample containing Ba(OH)_(2) is dissolved in 10 ml of 0.5 MHCl

1.	20g of sample containing Ba(OH)_(2) is dissolved in 10 ml of 0.5 MHCl 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)
Answer» Solution :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 %