Question

Experiment: Sodium Content in butter

Intro: An experiment conducted to determine that amount of sodium in butter through back titration of measuring chloride ions. the chloride ions themselves are not directly titrated ; rather a known excess of silver nitrate is added to the sample. A silver chloride precipitate immediately forms leaving behind the excess silver ions. the leftover silver ions are then titrated with ammonium thiocyanate forming a silver thiocyanate precipitate. Based on the amount of ammonium thiocyanate consumed, one can back calculate the amount of excess silver that remained after the formation of the silver chloride precipitate. since one knows how much silver nitrate was added originally, one can then calculate the amount of silver chloride which was precipitated and therefore the amount of chloride present. IN this experiment a 1.4897 mass of butter was placed in a 250 Erlenmeyer flask, then hated to melt the butter. once the butter melted 10 ml of water, 25ml of 0.05 M silver nitrate, and 10 ml of 2.00 M nitric acid was added. After boiling again for 10 minutes 60ml of water as added, 1.0 ml of iron (iii) indicator and 1.0 ml of chloroform was added. The samples were then titrated with 0.04M ammonium thiocyanate until the endpoint of 21.60 ml was reached.

Questions:

1. what was the thiocyanate consumed in mmol

2. excess silver nitrate in mmol

3. the chloride content in mmol

4. sodium content in mg

Equations:

Ag+ (aq) + Cl- (aq) = AgCl (s)

Ag + (aq) + SCN- (aq) = AgSCN (s)

Answer 1

1. What was the thiocyanate consumed in mmol

Number of moles of thiocyanate taken = Concentration (M) x Volume (L)

= 0.04 (mol L^-1) x 0.0216 L

= 8.64 x 10^-4 moles (or) 0.864 mmol

2. Excess silver nitrate in mmol

Excess silver nitrate is equal to the amount of ammonium thiocyanate consumed during the titration i.e., 0.864 mmol.

3. The chloride content in mmol

First calculate the number of moles of silver nitrate taken,

Number of moles of silver nitrate taken = Concentration (M) x Volume (L)

= 0.05 (mol L^-1) x 0.025 L

= 1.25 x 10^-3 mol or 1.25 mmol

The chlorine content present = Total silver nitrate taken - Excess silver nitrate

= 1.25 mmol - 0.864 mmol

= 0.386 mmol

4. Sodium content in mg

Amount of chlorine present is equal to the amount of sodium present.

mass of sodium = moles x molar mass

= 0.386 mmol x 23 (value taken from periodic table)

= 8.878 mg