You run a titration experiment to determine the mass of ascorbic acid (MM=176.12

Question

You run a titration experiment to determine the mass of ascorbic acid (MM=176.12 g/mole) in a vitamin C tablet. The tablet was dissolved in 100mL of water and the titration required 23.0 mL of 0.5M NaOH for complete neutralization of the acid, according to the following reaction: AscH2 (aq) + 2NaOH (aq) -> Asc^2- (aq) + 2H2O (l) + 2Na^+ (aq) a) Calculate the moles of NaOH used. b) Calculate the moles of acid in the solution. c) Calculate the mass of ascorbic acid in the tablet. d) Show how you prepared 500mL of the 0.5 M NaOH solution from a 10M NaOH stock solution.

Explanation / Answer

The chemical equation of reaction between Ascorbic acid and sodium hydroxide is shown below.

AscH2 (aq) + 2NaOH (aq) ----------> Asc2- (aq) + 2H2O (l) + 2Na+ (aq)

1 mole of ascorbic acid needs 2 moles of sodium hydroxide for complete neutralisation.

Calculation :

(molarity x volume )acid = (molarity x volume )base

Calculation of NaOH solution

Molarity of solution Mass of NaoH added Volume of solution

1M 40g (1 mole solute) 1000 mL ( 1 dm3 solution)

0.5M 20g (0.5 mole solute) 1000 mL

Molarity of solution and mass of solute are directly proportional.

X = (0.5M X 40g) / 1M

X = 20g

Molarity of solution Mass of NaoH added Volume of solution

1M 40g (1 mole solute) 1000 mL ( 1 dm3 solution)

0.5M 20 g 1000 mL

0.5M y 23 mL

Volume of the solution and mass of solute are directly proportional for fixed concentration of solution.

y = (20g x 23mL) / 1000 mL

y = 0.46 g

mass of NaoH present in 0.5 M, 23 mL sodium hydroxide solution is 0.46g.

Number of moles of NaOH = Mass of NaOH (g )    23Na ,16O , 1H

gram molecular mass of NaOH (g)

= 0.46 g / 40g

Number of moles of NaOH in 23 mL of solvent = 0.0115 mol

b) (Molarity x Volume)acid = (Molarity x Volume)base

Macid x 100mL = 0.5M x 23mL

Molarity of ascorbic acid solution = (0.5M x 23mL ) / 100 mL

Molarity of ascorbic acid solution = 0.115 M

Molarity Mass of ascorbic acid Volume of solution

1M 176.12g (1 mole solute) 1000 mL

10M 176.12g 100mL

0.115 M X 100 mL

Number of moles of ascorbic acid present in 100mL of solution = 0.115M

c) x = (0.115M x 176.12g )/ 10M

X = 2.025g This is the amount of ascorbic acid present in 100 mL of solution of tablet.

d) If 1 mole of solute is dissolved in 1000mL solution then the solution is 1 molar (M)

Weight of NaOH required = Molarity x Molecular mass x Volume of solution

1000

= 0.5 M x 40g x 500mL

1000mL

= 10g.

10g of NaOH is dissolved in water to get 500 mL solution of molarity 0.5M.

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