Q. Following above procedures, in standardization of AgNO3 I got volume of AgNO3

Question

Q. Following above procedures, in standardization of AgNO3 I got volume of AgNO3, after titration, is 19.9 ml for mass of NaCl 0.0580g and volume of AgNO3 20.5 ml for mass of NaCl 0.0595 g.

In 2nd procedure (determination of the percent halide) I got volume of AgNO3, after titration, is 26.7 ml in all three trials. Now the question is, Calculate and report the percent and the identity of the halide for each sample that was titrated. NOTE: In an unknown sample, mass of NaX +impurity in 50 mL of sample =0.1g; imputiry < or = 30% mass.

Explanation / Answer

AgNO3 standardization

Trial 1 : moles of NaCl = 0.0580 g/58.44 g/mol = 9.92 x 10^-4 M

moles of AgNO3 required = 9.92 x 10^-4 mols for neutralization

Molarity of AgNO3 solution = 9.92 x 10^-4 mols/0.0199 L = 0.05 M

Trial 2 : moles of NaCl = 0.0595 g/58.44 g/mol = 1.02 x 10^-3 M

moles of AgNO3 required = 1.02 x 10^-3 mols for neutralization

Molarity of AgNO3 solution = 1.02 x 10^-3 mols/0.0205 L = 0.05 M

Final concentration by averaging two reading of AgNO3 = 0.05 M

Identification of halide

moles of AgNO3 used = 0.05 M x 0.0267 L = 1.33 x 10^-3 mols

mass of NaX + impurity = 0.1 g/1.33 x 10^-3 mols = 75 g/mol

subtracting 30% of it = 75 - 22 = 53 g/mol

Which is close to the molar mass of NaCl

So the halide is Cl

percant of halide in sample = = 1.33 x 10^-3 x 35.4 x 100/0.1 = 47.08%

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