Answer :
Final answer:
To calculate the molarity of the H2SO4 solution, the number of moles of KOH reacted is first calculated, then the stoichiometry of the reaction is used to find the moles of H2SO4, which is then used to calculate the molarity of H2SO4. The calculated molarity does not match any of the provided options, indicating a potential error.
Explanation:
The question asks what the molarity of an H2SO4 solution is given that 18.6 mL of it neutralizes 30.5 mL of 1.55 M KOH solution.
To find the molarity of H2SO4, we use the stoichiometry of the acid-base neutralization reaction, which is given as 2 KOH + H2SO4
arr K2SO4 + 2 H2O, meaning that each mole of H2SO4 reacts with two moles of KOH.
First, we calculate the number of moles of KOH:
[0.0305 L KOH] x [1.55 mol KOH / L KOH] = 0.047275 moles KOH
Then we use the stoichiometry of the reaction where 2 moles of KOH react with 1 mole of H2SO4:
[0.047275 mol KOH] x [1 mol H2SO4 / 2 mol KOH] = 0.0236375 moles H2SO4
Finally, we find the molarity of the H2SO4 by dividing the moles of H2SO4 by the volume (in liters) of the H2SO4 solution:
[0.0236375 moles H2SO4 / 0.0186 L H2SO4] = 1.27 M
The correct answer is not listed among the options provided. There may be a mistake in either the provided answers or in the details of the question.
