Question

Chloride (35.45 g/mol) in a brine solution is determined by the Volhard method. A 14.17-ml aliquot of the solution is treated with 14.68 mL of standard 0.1009 M AgNO3 solution. The excess silver is titrated with standard 0.103 M KSCN solution, requiring 2.9 ml. to reach the red Fe(CN)2end point. Calculate the % (wv) Cr content of the solution (1) Agar Agai (2) AgSCN + AgSCN Note: Carry out the complete solution then round off the answer to 3 significant figures.

Answer 1

Volume of ce solution = 14.17 me Molarity of AgNO3 - O1009 M Volume of AgNO3 = 14.68 ml SCN= Molarity of 0-103 M Volume of SCN 2.9 me Required Millimoles: Molarity * Volume Millimo les of SCN 0-103 x 2.9 0-8987 mmol Millimoles of Agt added a Molarity & Volume 0.1009 X 14.68 1.4812 mmol Millimoles of Agt Millimoles of Agt Millimoles of Agt added Reacted with ce- Reacted with SON 1-4812 0-2987 1. 1825 2 mmol Millimales Volume a of Molalty х 14.17 1-1895- Molarity > 0.0835 M Molality of a

To calculate Concentration in (w/v) form Volume solution = loo me of a Mass . 35-5 g/mol Molai Molality: Mass Volume of solution Molal Mass 0.0835- Mass X looo 35.5 Too X Mass 0.296 g / 100 me % (w/w) of ce - 0-296 g/100 me