Calculation of weight of 90% pure sulphuric acid required to neutralize 5g of caustic soda

Introduction
Neutralization is a chemical reaction in which an acid and a base react with each other to form a salt and water. In this process, the pH of the solution is neutralized. Sulphuric acid is a strong acid and caustic soda is a strong base. When these two are mixed, they react to form a salt and water.

Formula used
The balanced equation for the reaction between sulphuric acid and caustic soda is as follows:
H2SO4 + 2NaOH → Na2SO4 + 2H2O

Calculation
To calculate the weight of 90% pure sulphuric acid required to neutralize 5g of caustic soda, we need to first calculate the amount of caustic soda and then use stoichiometric calculations to determine the amount of sulphuric acid required.

1. Calculation of moles of caustic soda
The molar mass of NaOH is 40 g/mol (23 for Na and 17 for O and H). Therefore, 5g of NaOH is equivalent to 5/40 = 0.125 moles of NaOH.

2. Calculation of moles of sulphuric acid
From the balanced equation, we know that 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, the amount of H2SO4 required to neutralize 0.125 moles of NaOH is 0.125/2 = 0.0625 moles.

3. Calculation of weight of sulphuric acid
The molar mass of H2SO4 is 98 g/mol (2 for H, 32 for S and 64 for O). Therefore, the weight of 0.0625 moles of H2SO4 is 0.0625 x 98 = 6.125 g.

However, we need to remember that the sulphuric acid mentioned in the question is 90% pure. Therefore, we need to adjust the weight of sulphuric acid accordingly.

4. Calculation of weight of 90% pure sulphuric acid
The weight of 100% pure H2SO4 required is 6.125/0.9 = 6.806 g. Therefore, the weight of 90% pure H2SO4 required to neutralize 5g of NaOH is 6.806 g.

Conclusion
In conclusion, 6.806 g of 90% pure sulphuric acid is required to neutralize 5g of caustic soda.