15.00 mL of 0.425 M H₂SO₄ is required to completely neutralize 23.9 mL of KOH. What is the molarity of the KOH?
2 KOH(aq) + H₂SO₄(aq) $\rightarrow$ K₂SO₄(aq) + 2 H₂O (l)

Question

15.00 mL of 0.425 M H₂SO₄ is required to completely neutralize 23.9 mL of KOH. What is the molarity of the KOH? 2 KOH(aq) + H₂SO₄(aq) $\rightarrow$K₂SO₄(aq) + 2 H₂O (l)

Quizplus · Accepted Answer

The balanced chemical equation shows that 2 moles of KOH react with 1 mole of the acid. First, calculate moles of acid: $0.425 \, \text{M} \times 0.015 \, \text{L} = 0.006375 \, \text{moles}$. Since the stoichiometry is 2:1, the moles of KOH required are $2 \times 0.006375 = 0.01275 \, \text{moles}$. To find the molarity of KOH: $0.01275 \, \text{moles} / 0.0239 \, \text{L} = 0.533 \, \text{M}$.

1500 ML of 0 →\rightarrow→ K2SO4(aq) + 2 H2O (L) A) 1

1500 ML of 0 $\rightarrow$ K₂SO₄(aq) + 2 H₂O (L)
A) 1