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

Answer:

The concentration of NaCl = 0.3374 M

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.

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

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

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Thus,  

$Moles\ of\ AgNO_3 =Molarity \times {Volume\ of\ the\ solution}$

$Moles\ of\ AgNO_3 =0.2503 \times {20.22\times 10^{-3}}\ moles$

$Moles\ of\ AgNO_3 = 5.0611 \times 10^{-3} moles$

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:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity\ of\ NaCl=\frac{5.0611\times 10^{-3}}{15.00\times 10^{-3}}$

$Molarity\ of\ NaCl= 0.3374 M$

Thus, the concentration of NaCl = 0.3374 M