Question

A 25.00 mL sample of of a sulfurous acid solution is completely neutralized using 17.12 mL of 0.13 M sodium hydroxide. What is the initial concentration of sulfurous acid?

Answer 1

Answer:

0.0445 M is the initial concentration of sulfurous acid.

Explanation:

$H_2SO_3+2NaOH\rightarrow Na_2SO_3+2H_2O$

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_3$

are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?\\V_1=25.00mL\\n_2=1\\M_2=0.13 M\\V_2=17.12 mL$

Putting values in above equation, we get:

$2\times M_1\times 25.00 mL=1\times 0.13 M\times 17.12 mL$

$x=\frac{1\times 0.13 M\times 17.12 mL}{2\times 25.00}=0.0445 M$

0.0445 M is the initial concentration of sulfurous acid.