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)

a) 0.125 M
b) 0.150 M
c) 0.175 M
d) 0.200 M

Answer 1

To determine the concentration of NaCl in the solution, the titration data is used in the formula M1V1 = M2V2. After calculation, the molarity of NaCl appears to be 0.335 M, but this does not match any of the provided options, indicating a need to reassess the calculations.

Step-by-step explanation:

We are looking to find the concentration of NaCl in the solution using a titration method. The reaction between NaCl and AgNO₃ is a one-to-one mole ratio according to the equation: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq).

To calculate the concentration of NaCl, we use the formula M1V1 = M2V2, where M1 is the molarity of AgNO₃, V1 is the volume of AgNO₃ used, M2 is the molarity of NaCl, and V2 is the volume of NaCl solution. Plugging in the given values:

M1 = 0.2503 M (AgNO₃)
V1 = 20.22 mL (AgNO₃)
V2 = 15.00 mL (NaCl)

We're solving for M2 (the concentration of NaCl), so rearrange the formula to solve for M2:

M2 = (M1V1) / V2 = (0.2503 M * 20.22 mL) / 15.00 mL

After performing the calculations, we find that the molarity of NaCl is 0.335 M, which is not listed in the options provided.

Therefore, we need to double-check the calculations and ensure there are no mistakes, as none of the options (a) 0.125 M, (b) 0.150 M, (c) 0.175 M, or (d) 0.200 M match the calculated value.