If it takes 10.0 mL of 2.0 M H 2SO4 to neutralize 30.0 mL of KOH, what is the molarity of KOH?
To determine the molarity of 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 1 mole of H2SO4 reacts with 2 moles of KOH. Therefore, the number of moles of KOH is half of the number of moles of H2SO4 used in the reaction.
First, let's calculate the number of moles of H2SO4 used:
Moles of H2SO4 = Molarity (mol/L) x Volume (L)
Moles of H2SO4 = 2.0 mol/L x 0.010 L
Moles of H2SO4 = 0.020 mol
Since the molarity of KOH is unknown, let's denote it as x mol/L. The number of moles of KOH is equal to the number of moles of H2SO4 used:
Moles of KOH = Moles of H2SO4
x mol/L x 0.030 L = 0.020 mol
x = 0.020 mol / 0.030 L
x = 0.67 mol/L
Therefore, the molarity of KOH is 0.67 M.