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------------------------------------------------ What is the molarity of the [tex]$H_2SO_4$[/tex] solution, 18.6 mL of which neutralizes 30.5 mL of a 1.55 M [tex]$KOH$[/tex] solution?

A. 0.47 M
B. 1.55 M
C. 0.96 M
D. 0.31 M

The molarity of the (H₂SO₄) solution, (18.6, mL) of which neutralizes (30.5, mL) of (1.55, M) (KOH) solution is 1.27 M, hence the correct answer is none of the above options.

Explanation:

To find the molarity of the H2SO4 solution that neutralizes a given volume of KOH solution, we can use the stoichiometry of the reaction and the molarity of the KOH solution. The balanced chemical equation for the reaction between H2SO4 and KOH is:

2 KOH + H2SO4 -> K2SO4 + 2 H2O

According to the stoichiometry of the reaction, 2 moles of KOH react with 1 mole of H2SO4. We can calculate the moles of KOH that reacted as follows:

moles of KOH = volume of KOH x molarity of KOH = 30.5 mL x 1.55 mol/L = 0.047275 mol (Note: convert milliliters to liters by dividing by 1000)

Since 2 moles of KOH react with 1 mole of H2SO4, the moles of H2SO4 that reacted are half that of KOH:

moles of H2SO4 = moles of KOH / 2 = 0.047275 mol / 2 = 0.0236375 mol

To find the molarity of the H2SO4 solution, divide the moles of H2SO4 by the volume of the H2SO4 solution in liters:

molarity of H2SO4 = moles of H2SO4 / volume of H2SO4 = 0.0236375 mol / 0.0186 L = 1.27 M