What volume of a 0.150 M solution of the strong base KOH must be added to completely neutralize 200.0 mL of a 0.120 M solution of H₂SO₄?

Question

| General Chemistry 2

Accepted Answer

Equation: 2 KOH (aq) + H₂SO₄ (aq) → K₂SO₄ (aq) + 2 H₂O (l)

2 mole of KOH is required to neutralized 1 mole of H₂SO₄

⇒ when the endpoint of the neutralization is reached: n_KOH = 2 x n_H2SO4

⇒ M_KOH x V_KOH = 2 x M_H2SO4 x V_H2SO4

⇒ V_KOH = $\frac{2 \times M_{H 2 S O 4} \times V_{H 2 SO 4}}{M_{K O H}}$

⇒ V_KOH = $\frac{2 \times 0.120 \times 200.0}{0.150}$ = 320 mL

Midterm 1 - Question 6 | General Chemistry 2