If 91.9 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.684 g of precipitate, what is the molarity of silver ion in the original solution?

0.052M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

AgNO3 (aq) + KCl (aq) → AgCl (s) + KNO3 (aq)

Next, we shall determine the number of mole present in 0.684 g of precipitate, AgCl. This is illustrated below:

Mass of AgCl = 0.684g

Molar mass of AgCl = 108 + 35.5 = 143.5g/mol

Number of mole of AgCl =.?

Mole = Mass / Molar Mass

Number of mole of AgCl = 0.684/143.5

Number of mole of AgCl = 4.77*10¯³ mol

Therefore, 4.77*10¯³ mol of AgCl was produced.

Next, we shall determine the number of mole AgNO3 that reacted. This can be achieved by doing the following:

From the balanced equation above,

1 mole of AgNO3 reacted to produce 1 mole of AgCl.

Therefore, 4.77*10¯³ mol of AgNO3 will also react to produce 4.77*10¯³ mol of AgCl.

Next, we shall determine the molarity of AgNO3. This can obtain as follow:

Volume = 91.9mL = 91.9/1000 = 0.0919L

Mole of AgNO3 = 4.77*10¯³ mol

Molarity =.?

Molarity = mole / Volume

Molarity = 4.77*10¯³ / 0.0919

Molarity = 0.052M.

Therefore, the molarity of the AgNO3 is 0.052M.

Finally, we shall determine the molarity of silver ion in AgNO3. This can be obtained as follow:

AgNO3 (aq) - > Ag + (aq) + NO3 - (aq)

From the balanced equation above,

1 mole of AgNO3 produced 1 mole of Ag+.

Therefore, 0.052M AgNO3 will also produce 0.052M Ag+

Therefore, the molarity of the silver ion, Ag + is 0.052M