Given AgNO 3 (aq) + NaCl(aq)→ AgCl(s) + NaNO 3 (aq), if titration of 30.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL, of NaCl to reach the end point, what is the concentration of NaCl in the solution?

0.3714 M

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

AgNO3 (aq) + NaCl (aq) -> AgCl(s) +

NaNO3 (aq)

Step 2:

Determination of the number of moles of AgNO3 in 30.00 mL of 0.2503 M AgNO3. This is illustrated below:

Molarity of AgNO3 = 0.2503 M

Volume of solution = 30mL = 30/1000 = 0.03L

Number of mole of AgNO3 =?

Molarity = mole /Volume

0.2503 = mole / 0.03

Cross multiply to express in linear form

Mole = 0.2503 x 0.03

Mole of AgNO3 = 7.509x10^-3 mole

Step 3:

Determination of the number of mole of NaCl required to react with AgNO3. This is illustrated below:

AgNO3 (aq) + NaCl (aq) -> AgCl(s) +

NaNO3 (aq)

From the balanced equation above,

1 mole of AgNO3 required 1 mole of NaCl for complete neutralization.

Therefore, 7.509x10^-3 mole of AgNO3 will also require 7.509x10^-3 mole of NaCl for complete neutralization.

Step 4:

Determination of the molarity of NaCl. This is illustrated below:

Mole of NaCl = 7.509x10^-3 mole

Volume of solution = 20.22 mL = 20.22/1000 = 0.02022L

Molarity =?

Molarity = mole /Volume

Molarity of NaCl = 7.509x10^-3/0.02022

Molarity of NaCl = 0.3714 M

The concentration of NaCl in the solution is 0.371 M

Explanation:

Step 1: Data given

Volume of 0.2503 M AgNO3 solution = 30.00 mL = 0.030 L

Volume of NaCl = 20.22 mL = 0.02022 L

Step 2: The balanced equation

AgNO3(aq) + NaCl(aq) ===> AgCl(s)  + NaNO3(aq)

Step 3: Calculate concentration of NaCl

C1*V1 = C2*V2

⇒with C1 = the concentration of AgNO3 = 0.2503 M

⇒with V1= the volume of AgNO3 = 0.030 L

⇒with C2 = the concentration of NaCl = TO BE DETERMINED

⇒with V2 = the volume of NaCl = 0.02022 L

0.2503 M * 0.030 L = C2 * 0.02022L

C2 = (0.2503 * 0.030) / 0.02022

C2 = 0.371 M

The concentration of NaCl in the solution is 0.371 M

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