Question

You have 3.00 L of a 3.00 M solution of NaCl(aq) called solution A. You also have 2.00 L of a 2.00 M solution of AgNO3(aq) called solution B. You mix these solutions together, making solution C. Hint: AgCl is a precipitate. Calculate the concentrations (in M) of the following ions in solution C. NO3-

Answer 1

Answer:

The concentration of nitrate ions in solution is 0.8 M.

Explanation:

In solution A;

$NaCl(aq)\rightarrow Na^+(aq)+Cl^-(aq)$

Moles of sodium chloride in solution:

$3.00M=\frac{moles}{3.00L}$

Moles of sodium chloride = 9 mole

In solution B;

$AgNO_3(aq)\rightarrow Ag^+(aq)+NO_{3}^-(aq)$

Moles of sodium chloride in solution:

$2.00M=\frac{moles}{2.00L}$

Moles of sodium chloride = 4 mole

In solution C;

$AgNO_3+NaCl\rightarrow AgCl+NaNO_3$

1 mol of silver chloride reacts with 1 mol of sodium chloride to give 1 mol of silver chloride and 1 mol of sodium nitrate.

Then 4 mol silver chloride reacts with 4 mol of sodium chloride to give 4 mol of silver chloride and 4 mol of sodium nitrate.

1 mol silver nitrate gives 1 mol of nitrate ions.

So, 4 moles of silver nitrate will gives 4 mole of nitrate.

Total volume of solution = 3.00 L+ 2.00 L = 5.00 L

Concentration of nitrate ions in solution:

$[NO_{3}^-]=\frac{4 mol}{5 L}=0.8 mol/L$

The concentration of nitrate ions in solution is 0.8 M.