An acid solution is 0.110 M in HCl and 0.210 M in H2SO4. What volume of a 0.150 M solution of KOH must be added to 550.0 mL of the acidic solution to completely neutralize all of the acid?
Answer
Neutralizing a Mixed Acid Solution with Potassium Hydroxide (KOH)
🧪 Given:
Volume of acid solution = 550.0 mL = 0.550 L
[HCl] = 0.110 M
[H2SO4] = 0.210 M
[KOH] = 0.150 M
[HCl] = 0.110 M
[H2SO4] = 0.210 M
[KOH] = 0.150 M
📘 Step 1: Calculate Moles of Acidic Protons
HCl is a strong monoprotic acid (1 H+ per molecule):
Moles of HCl = 0.110 mol/L × 0.550 L = 0.0605 mol
H2SO4 is a strong diprotic acid (2 H+ per molecule):
Moles of H2SO4 = 0.210 mol/L × 0.550 L = 0.1155 mol
Total H+ from H2SO4 = 2 × 0.1155 = 0.231 mol
Total H+ from H2SO4 = 2 × 0.1155 = 0.231 mol
➕ Step 2: Total Moles of H+ (Acid)
Total moles of H+ = 0.0605 + 0.231 = 0.2915 mol
🧪 Step 3: Neutralization Reaction
Each mole of KOH provides 1 mole of OH−, which neutralizes 1 mole of H+:
KOH + H+ → H2O + K+
So, moles of KOH required = moles of H+ = 0.2915 mol
🧮 Step 4: Calculate Volume of KOH Required
Moles = Molarity × Volume
Volume = Moles / Molarity = 0.2915 mol / 0.150 mol/L ≈ 1.943 L
Volume = Moles / Molarity = 0.2915 mol / 0.150 mol/L ≈ 1.943 L
✅ Final Answer:
To completely neutralize the acidic solution, 1.943 liters (or 1943 mL) of 0.150 M KOH must be added.
Note: This calculation assumes complete dissociation and a one-to-one neutralization of acid and base equivalents.