# 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?

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###### 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?

## Answers

Answer:0.0445 M is the initial concentration of sulfurous acid.Explanation:[tex]H_2SO_3+2NaOH\rightarrow Na_2SO_3+2H_2O[/tex]To calculate the concentration of acid, we use the equation given by neutralization reaction:[tex]n_1M_1V_1=n_2M_2V_2[/tex]where,[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_3[/tex] are the n-factor, molarity and volume of base which is NaOH.We are given:[tex]n_1=2\\M_1=?\\V_1=25.00mL\\n_2=1\\M_2=0.13 M\\V_2=17.12 mL[/tex]Putting values in above equation, we get:[tex]2\times M_1\times 25.00 mL=1\times 0.13 M\times 17.12 mL[/tex][tex]x=\frac{1\times 0.13 M\times 17.12 mL}{2\times 25.00}=0.0445 M[/tex]0.0445 M is the initial concentration of sulfurous acid.

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