Question

What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL of the AgNO3 solution to reach the end point? AgNO3 (aq) + NaCl(aq) ⟶ AgCl(s) + NaNO3 (aq)

Answer 1

The concentration of NaCl = 0.3374 M

Step-by-step explanation:

Given :

Molarity of AgNO₃ = 0.2503 M

Volume of AgNO₃ = 20.22 mL

The conversion of mL into L is shown below:

`1 mL= 10^(-3) L`

Thus, volume of the solution = 20.22×10⁻³ L

Molarity of a solution is the number of moles of solute present in 1 L of the solution.

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

The formula can be written for the calculation of moles as:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

Thus,

`Moles\ of\ AgNO_3 =Molarity * {Volume\ of\ the\ solution}`

`Moles\ of\ AgNO_3 =0.2503 * {20.22* 10^(-3)}\ moles`

`Moles\ of\ AgNO_3 = 5.0611 * 10^(-3) moles`

The chemical reaction taking place:

`AgNO_3_(aq) + NaCl_(aq) \rightarrow AgCl_(s) + NaNO_3_(aq)`

According to reaction stoichiometry:

1 mole of AgNO₃ reacts with 1 mole of NaCl

Thus,

5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl

Thus, moles of NaCl required = 5.0611×10⁻³ moles

Volume of NaCl required = 15.00 mL

The conversion of mL into L is shown below:

`1 mL= 10^(-3) L`

Thus, volume of the solution = 15.00×10⁻³ L

Applying in the formula of molarity as:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

`Molarity\ of\ NaCl=(5.0611* 10^(-3))/(15.00* 10^(-3))`

`Molarity\ of\ NaCl= 0.3374 M`

Thus, the concentration of NaCl = 0.3374 M