Question

Equal volumes of 0.1 M AgMO3 and 0.2M NaCl are mixed. The concentration of NO³- ions in the mixture will be??

Answer 1

Let the volume of both the solution is 1 L.

Total volume of the mixture after mixing = 2L

Number of moles in 1L of NaCl solution = 0.2

Number of moles in 1L of AgNO3 solution = 0.1

In solution 

[NO3-]= [AgNO3]= 0.1

 

After mixing

[NO3-] = 0.1/2

0.05M

Answer 2

Answer:

The concentration of $NO_3^-$ions in the mixture will be 0.05 mol/L.

Explanation:

$AgNO_3(aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3(aq)$

1 mol of silver nitrate reacts with 1 mol of sodium chloride

So, 0.1 M of silver nitrate will completely reacts with 0.1 M of sodium chloride.

0.1 M of silver nitrate means that 0.1 mol of silver nitrate in 1L . 1 mol of silver nitrate gives 0.1 mole of nitrate ions in 1L solution.

After mixing the volume of the solution = 2 L

Molarity of nitrate in 2 L solution is:$\frac{0.1 mol}{2L}=0.05 mol/L$

The concentration of $NO_3^-$ions in the mixture will be 0.05 mol/L.