Question

First, he standardized the AgNO₃ titrant with 0.1242 g dry NaCl dissolved in 25.0 ml DI water, and 26.25 ml of titrant was consumed. Then he used the standardized titrant to determine the concentration of chloride in the solution. 11.71 ml of AgNO₃ solution was used to titrate 25.00 ml of solution

a) Calculate the concentration (M) of AgNO₃ titrant.

b) Determine the content of chloride in the unknown solution (M).

Answer 1

one g equivalent of AgNO3 reacts with 1 g equivalent of NaCl

normality of NaCl = 0.1242X 1000 / 58.5 X 25 = 0.0849 N

G equvalents of NaCl = 0.0849 X 25 = 2.1225 milligm equilvalents

so g equivalents of AgNO3 = 2.1225 g equivalents

so normality / molarity of AgNO3 = 2.1225 / 26.25( volume) = 0.0809 N / M

2)

N1V1 = N2V2   ( milliequivalent of titrant = milliequivalents of Cl-)

0.0809 X 11.71 = 25 X Normality of Cl-

normality / molarity of Chloride ions = 0.0378 N /M