Q80E - Chemistry (OpenStax)

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)

Step-by-Step Solution

Answer

0.3374 M 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.

Given data

This titration is between acid and base. For this formula we use 

\({M_a}{V_a} = {M_b}{V_b}\)

    

Molarity of AgNO₃ =   0.2503 M       and     Volume of Ag NO₃ = 20.22 mL = 0.02022 L

 

Molarity of AgNO₃ = M X L = 0.2503 x 0.02022 = 5.061 x 10 ^-3 mol of Ag NO3

 

Molarity of NaCl   =   ?        Volume of NaCl = 15 mL =  0.0150 L

Determine the concentration

AgNO₃(aq) + NaCl(aq) ⟶ AgCl(s) + NaNO₃(aq)

\(\begin{aligned}{\underline{\phantom{xx}}}{M_a}{V_a} = {M_b}{V_b}\\5.061{\rm{ }}x{\rm{ }}{10^{ - 3}}mol{\rm{ }}of{\rm{ }}Ag{\rm{ }}N{O_3}\; \times \frac{{1\,mol\,NaCl}}{{1mol{\rm{ }}of{\rm{ }}Ag{\rm{ }}N{O_3}\;}} = 5.061{\rm{ }}x{\rm{ }}{10^{ - 3}}mol{\rm{ }}\;\end{aligned}\)

Concentration of NaCl 

= \(\begin{aligned}{\underline{\phantom{xx}}}\frac{{5.061{\rm{ }}x{\rm{ }}{{10}^{ - 3}}mol{\rm{ }}}}{{0.0150L}}\\ = 0.3374\,M\,of\,NaCl\end{aligned}\)