What mass of KOH is required to prepare 500 ~mL of an aqueous stock solution, if 20 ~mL of this solution exactly neutralizes 30 ~mL of 0.1 M HNO ₃ ? (Molar mass of KOH = 56 ~g ~mol ⁻¹ )

Step-by-step solution

For the neutralization reaction between KOH and HNO ₃ : M ₁ V ₁ = M ₂ V ₂ M _ KOH 20 = 0.1 30 M _ KOH = 3 20 = 0.15 M Now, calculate the mass of KOH required for 500 ~mL ( 0.5 ~L ) of this solution: Molarity = Mass Molar mass Volume (L) 0.15 = w 56 0.5 w = 0.15 56 0.5 = 4.2 ~g Answer: 4.2 ~g