Question

The volume of sulfuric acid, H2SO4 0.15 mol dm^-3 required to neutralize 25.0 cm^3 of potassium hydroxide solution, KOH in full is 30.5 cm^3. Calculate the molarity of potassium hydroxide solution, KOH.

a) 0.12 mol dm^-3
b) 0.15 mol dm^-3
c) 0.18 mol dm^-3
d) 0.20 mol dm^-3

Answer 1

The molarity of the potassium hydroxide solution, KOH, is approximately 0.183 mol/dm³, which is closest to option c) 0.18 mol/dm³.

Step-by-step explanation:

To calculate the molarity of the potassium hydroxide solution (KOH), we need to use the balanced chemical equation for the reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH):

H2SO4 + 2KOH -> K2SO4 + 2H2O

From the equation, we can see that one mole of H2SO4 reacts with 2 moles of KOH to form K2SO4 and 2 moles of water.

Given that the volume of H2SO4 required to neutralize 25.0 cm3 of KOH in full is 30.5 cm3, we can set up the following equation to solve for the molarity of KOH:

(0.15 mol/dm3)(30.5 cm3) = x (mol/dm3)(25.0 cm3)

Simplifying the equation, we find that the molarity of KOH is approximately 0.183 mol/dm3, which is closest to option c) 0.18 mol/dm3.