Question

A 12.72 g sample of silver ore was treated with nitric acid to dissolve the silver. After filtering, it was made to a volume of 250mL. 25mL of this sample was treated with 50mL of 0.1011M potassium chloride, which precipitated all the silver and left chloride in excess. The excess chloride was titrated with 0.0871M silver nitrate solution, and required 15.69mL. Calculate the percent of silver in the ore

Answer 1

Reaction of silver with nitric acid :

3Ag(s)+4HNO3(aq) ---->  3AgNO3(l)+NO(g)+2H2O (l) (For dilute nitric acid)

Ag(s)+2HNO3(aq) ----> 4AgNO3(l)+ NO2(g)+H2O (l)   (For concentrated nitric acid)

Combined equation :

4Ag(s)+6HNO3(aq) ---->  3AgNO3(l)+NO(g) + NO2(g)+ 3H2O (l)

We have to find the percent of silver in ore which we can find out using the formula : (mass(m)Ag / total mass)*100

total mass of sample = 12.72g (given in question)

we have to find out m(Ag) which we can find out by using formula:

m(Ag) = n(Ag) * molar mass of (Ag) (molar mass of Ag = 107.87g/mol)

and for finding n(Ag) we have to follow the following steps:

1) In the question it is given that excess of chloride was titrated with silver nitrate solution

Therefore, n(no. of moles) Cl^- (in excess) = n AgNO₃

Molarity = no. of moles(n) / volume of solution (L)

n AgNO₃ = molarity(M) * volume of solution in L

Molaritiy(M) = 0.0871M Volume = 15.69ml = 15.69 *10^-3 L(this data is given in the question)

n AgNO₃ = 0.0871 * 15.69 *10^-3 = 1.367 * 10^-3 mol

since, n(no. of moles) Cl^- (in excess) = n AgNO₃

n(no. of moles) Cl^- (in excess) = 1.367 * 10^-3 mol of Cl- in excess

2) using the same method we can find out number of moles of Cl- in KCl solution

Molarity(M) = 0.1011M = volume = 50ml = 50*10^-3L

n(Cl-) in KCl solution = 0.101 * 50*10^-3 = 5.05*10^-3 mol of Cl-

3) Now we have to find out moles of silver(Ag) in 25mL of solution =  (n)Cl^- in KCl solution - (n)Cl^- in excess

n(Ag) in 25mL =  5.05*10^-3 - 1.367 * 10^-3 = 3.68*10^-3 mol of Ag

4) now, n(Ag) in 250mL = 3.68*10^-3 * 250/25 = 36.8 * 10^-3 mol of Ag

5) m(mass)(Ag) = n(Ag) * molar mass of Ag (molar mass of Ag = 107.87g/mol)

m(Ag) = 36.8 * 10^-3 * 107.87 = 3.969 g of Ag

% of Ag = 100* m(Ag) / m(total)

% of Ag = 100 * (3.969) / 12.72 = 31.20

% 0f Ag = 31.20%