If 97.7 ml of silver nitrate solution reacts with excess potassium chloride solution to yield 0.326 g of precipitate, what is the molarity of silver ion in the original solution?

Question

Chemistry

Cherish Barrera

3 September, 05:19

If 97.7 ml of silver nitrate solution reacts with excess potassium chloride solution to yield 0.326 g of precipitate, what is the molarity of silver ion in the original solution?

+1

Answers (1)

Know the Answer?

Lovely · Answer 1

The complete balanced chemical reaction is written as:

AgNO3 + KCl - - - > AgCl + KNO3

where AgCl is our precipitate

So calculating for moles of AgCl produced: MM AgCl = 143.5 g/mol

moles AgCl = 0.326 g / (143.5 g/mol) = 2.27 x 10^-3 mol

we see that there is 1 mole of Ag per 1 mole of AgCl so:

moles Ag = 2.27 x 10^-3 mol

The molarity is simply the ratio of number of moles over volume in Liters, therefore:

Molarity = 2.27 x 10^-3 mol / 0.0977 L

Molarity = 0.0233 M