In a titration process, 15 cm3 of 0.5 M dilute sulphuric acid is completely used to neutralise10 cm3 of sodium hydroxide solution.Calculate the molarity of sodium hydroxide solution with the help of a balanced chemicalequation. (6 Marks
Question
In a titration process, 15 cm3 of 0.5 M dilute sulphuric acid is completely used to neutralise10 cm3 of sodium hydroxide solution.Calculate the molarity of sodium hydroxide solution with the help of a balanced chemicalequation. (6 Marks
Solution
Sure, here is the step by step solution:
Step 1: Write the balanced chemical equation for the reaction. The reaction between sulphuric acid (H2SO4) and sodium hydroxide (NaOH) is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
From the balanced equation, we can see that one mole of sulphuric acid reacts with two moles of sodium hydroxide.
Step 2: Calculate the number of moles of sulphuric acid used in the reaction.
Number of moles = Molarity x Volume (in litres) Number of moles of H2SO4 = 0.5 M x 15 cm3 x (1/1000) = 0.0075 moles
Step 3: From the balanced equation, we know that 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, the number of moles of NaOH in the 10 cm3 solution is twice the number of moles of H2SO4, which is 2 x 0.0075 = 0.015 moles.
Step 4: Finally, calculate the molarity of the sodium hydroxide solution.
Molarity = Number of moles / Volume (in litres) Molarity of NaOH = 0.015 moles / 10 cm3 x (1/1000) = 1.5 M
So, the molarity of the sodium hydroxide solution is 1.5 M.
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