Three samples, claiming to each contain lead, weigh approximately 500mg each. Th

Question

Three samples, claiming to each contain lead, weigh approximately 500mg each. They are analyzed using the Mohr method.

A) Preparation of the titrant: Enough titrant for three standardizations and 3 analyses must me made. How many grams of silver nitrate must be weighed out in order to make 250 mL of 0.1 N titrant.

B) Standardization of the titrant: three pure samples of sodium chloride have been dried and dissolved into 50mL of water containing potassium dichromate indicator. Titration data for the silver nitrate solution you made is:

Run 1 - 270.5 mg NaCl, 43.32 mL AgNO3.

Run 2 - 286.2 mg NaCl, 45.70 mL AgNO3.

Run 3 - 280.6 mg NaCl, 44.88 mL AgNO3.

What is the concentration of the silver nitrate solution and error expressed in standard deviation units?

C) Each sample (of supposed lead) is dissolved into 50.0 mL of 0.5 M nitric acid. Sodium Chloride was added to the solution, precipitating PbCl2. Again, potassium dichromate was added as an indicator and the solution was titrated with 0.09876 N AgNO3.

Run 1 - 512.3 mg sample, 456.3 mg NaCl, 45.31 mL AgNO3.

Run 2 - 521.3 mg sample, 461.4 mg NaCl, 45.55 mL AgNO3.

Run 3- 531.2 mg sample, 449.7 mg NaCl, 42.86 mL AgNO3.

what is the weight percent of lead in the sample, and indicate error in units of standard deviation.

D) There is apparently a legal limit of 60 percent lead in the paint. Calculate the titration error based upon the composition of the solution at the endpoint:

Total Volume at endpoint- 100mL

Initial Volume of 0.05 N HNO3 - 50 mL

Mass NaCl added - 450mg

K2CrO4 added as an indicator- 0.2 mL of 1M indicator solution.

Explanation / Answer

For the titration

A) Preparation of titrant

For 250 ml of 0.1 M AgNO3 = 0.1 M x 0.250 L x 169.87 g/mol = 4.25 g AgNO4 needed

B) standarisation of titrant

Run 1 : molarity of NaCl = 0.2705 g/58.44 g/mol x 0.05 L = 0.0926 M

moles of NaCl = 0.0926 M x 0.05 L = 0.00463 mol

molarity of AgNO3 solution = 0.00463 mol/0.04332 L = 0.107 M

Likewise other runs can be calculated

C) Run 1

moles of NaCl initialy added = 0.4563 g/58.44 g/mol = 0.00781 mol

moles of AgNO3 added = 0.09876 M x 0.04531 ml = 0.0045 mol

moles of Cl reacted with Pb = 0.00781 - 0.0045 = 0.00331 mol

moles of PbCl2 = 0.00331/2 = 0.001655 mol

mass of Pb in sample = 0.001655 mol x 207.2 g/mol = 0.343 g

% Pb in sample = 0.343 x 100/0.5123 = 66.95%

Similarly other run can be calculated

D) Titration error for 6.95% Pb

Added excess Cl in solution = 0.0695 x 2/207.2 = 0.0007 mol

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