Answer: The volume of NaOH required is 10.5 mL
Explanation:
To calculate the molarity of solution, we use the equation:
\(\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\)
Mass of solute (sulfurous acid) = 0.0475 g
Molar mass of sulfurous acid = 82 g/mol
Volume of solution = 250. mL
Putting values in above equation, we get:
\(\text{Molarity of sulfurous acid}=\frac{0.0475g\times 1000}{82g/mol\times 250.mL}\\\\\text{Molarity of sulfurous acid}=2.32\times 10^{-3}M\)
- To calculate the volume of base, 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 \(H_2SO_3\)
\(n_2,M_2\text{ and }V_2\) are the n-factor, molarity and volume of base which is NaOH.
We are given:
\(n_1=2\\M_1=2.32\times 10^{-3}M\\V_1=250.mL\
_2=1\\M_2=0.1100M\\V_2=?mL\)
Putting values in above equation, we get:
\(2\times 2.32\times 10^{-3}\times 250.=1\times 0.1100\times V_2\\\\V_2=10.5mL\)
Hence, the volume of NaOH required is 10.5 mL
