Answer to Question #150864 in General Chemistry for Joe

Question

An impure sample of barium hydroxide of mass 1.3467 g was allowed to react with 50cm3 of 0.35 mol/dm3 HCL. When the excess acid was titrated against 0.25 mol/dm3NaOH, 8.5cm3 of NaOH solution was required.calculate the percentage purtiy of the sample of barium hydroxide

Answer 1

Total moles of HCl = Vol.of soln. * Conc.of soln.

= 0.05 * 0.35 moles

= 0.0175 moles

Excess acid left = NaOH used for titration

= Vol.of soln. * Conc.of soln.

= 0.0085 * 0.25 moles

= 0.002125 moles.

Moles used in reaction with Ba(OH)₂ = 0.015375

2HCl + Ba(OH)₂ $\to$ BaCl₂ + 2H₂O

Therefore, one mole of HCl neutralizes 0.5 moles of Ba(OH)₂

This implies 0.172875 moles neutralizes 0.0076875 moles

Mass of Ba(OH)₂ utilized = Molar mass * no. of moles

= 171.34 * 0.0076875 g

= 1.3172 g

Purity ratio = (1.3172/1.3467)

= 0.9781

Purity percent = Purity ratio * 100

= 97.81 %