A mixture sample of 3.50 g containing Ag salt and Pb salt in the form of nitrate

Question

A mixture sample of 3.50 g containing Ag salt and Pb salt in the form of nitrates (AgNO3 and Pb(NO3)2), to analyze their mass percentage, we dissolve the mixture in water and prepare a 500 ml solution. Then we perform the following two separate analyses:

Take 100 ml aliquot and add sufficient amount of Na2SO4 to form precipitates of Ag2SO4 and PbSO4, after cleaning, and dry out, the weight of precipitates is 0.577g

Take 200 ml aliquot and add sufficient amount of Na2S to form precipitates of Ag2S and PbS, after cleaning, and dry out, the weight of precipitates is 0.914g

MM (PbS)=239.3 g/mol; MM (Ag2S)=247.80 g/mol; MM (PbSO4)=303.26 g/mol;

MM (AgSO4)=311.80 g/mol; MM(AgNO3)=169.87 g/mol; MM (Pb(NO3)2)=331.28 g/mol

Find AgNO3 mass% and Pb(NO3)3 mass %.

Explanation / Answer

The total mass of mixture containing both the AgNO3 and Pb(NO3)2 = 3.5 g

The volume of the solution by dissolving the mixture in water = 500 mL

The volume of aliquot = 100 mL, i.e. 1/5th of the original solution

By the addition of sufficient amount of Na2SO4 to form precipitates of Ag2SO4 and PbSO4, after cleaning, and dry out, the mass of precipitates = 0.577g

i.e. For the whole solution, the mass of precipitates correspond to 5*0.577 = 2.885 g

The volume of aliquot = 200 mL, i.e. 2/5th of the original solution

By the addition of sufficient amount of Na2S to form precipitates of Ag2S and PbS, after cleaning, and dry out, the weight of precipitates = 0.914g

i.e. For the whole solution, the mass of precipitates correspond to (5/2)*0.914 = 2.285 g

The molar mass of PbS = 239.3 g/mol; that of Ag2S = 247.80 g/mol; that of PbSO4 = 303.26 g/mol;

that of Ag2SO4 = 311.80 g/mol; that of AgNO3 = 169.87 g/mol; that of Pb(NO3)2 = 331.28 g/mol

Here, you need to find out the mass%s of AgNO3 as well as Pb(NO3)3.

2AgNO3 + Na2S(O4) = Ag2S(O4) + 2NaNO3

Pb(NO3)2 + Na2S(O4) = PbS(O4) + 2NaNO3

i.e. The mixture of x moles of AgNO3 and y moles of Pb(NO3)2 (i.e. 3.5 g of mixture) needs (x/2 + y) moles of Na2S(O4) to form x/2 moles of Ag2S(O4) and y moles of PbS(O4) (i.e. 2.885 g of the corresponding precipitate) and (x + 2y) moles of NaNO3.

x*169.87 + y*331.28 + (x/2 + y)*142 = x/2 * 311.8 + y*303.26 + (x+2y)*85 ....... equation 1

And x*169.87 + y*331.28 + (x/2 + y)*78 = x/2 * 247.8 + y*239.3 + (x+2y)*85 ....... equation 2

By solving the above two equations, you will get both x and y values.

Now, the mass % of AgNO3 = (x*169.87/3.5)*100 and that of Pb(NO3)2 = (y*331.28/3.5)*100

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.