Question

An acid solution is 0.100 M in HCl and 0.210 M in H2SO4. What volume of a 0.150 M solution of KOH must be added to 500.0 mL of the acidic solution to completely neutralize all of the acid?

Answer 1

Answer:

When we add 0.33L of KOH only the HCl solution will be neutralized. When we add 1.4 L of KOH, all of the acid (the HCl and H2SO4 solution) will be neutralized.

Explanation:

Step 1: Data given

The solution has 0.100 M HCl and 0.210 M H2SO4

Molarity KOH = 0.150 M

Volume of acid solution = 500 mL = 0.5 L

Step 2: Calculate moles of HCl

Moles HCl = Molarity HCl * volume

Moles HCl = 0.100 M * 0.5 L

Moles HCl = 0.05 moles

Step 3: Calculate moles of H2SO4

Moles H2SO4 = 0.210 M * 0.5 L

Moles H2SO4 = 0.105 moles

Step 4: The balanced neutralization reaction of KOH with HCl can be written as:

KOH + HCl → KCl + H2O

The mole ratio is 1:1

This means to neutralize 0.05 moles HCl, we need 0.05 moles KOH

Step 5: The balanced neutralization reaction of KOH with H2SO4 can be written as:

2KOH + H2SO4 → K2SO4 + 2H2O

The mole ratio KOH: H2SO4 is 2:1

This means to neutralize 0.105 moles H2SO4 we need 0.210 moles KOH

Step 6: Calculate volume of KOH needed to neutralize the solution

To neutralize the HCl solution: 0.05 moles / 0.150 M = 0.33 L KOH needed

To neutralize the H2SO4 solution: 0.210 moles / 0.150 M = 1.4 L KOH needed

When we add 0.33L of KOH the HCl solution will be neutralized. When we add 1.4 L of KOH the HCl and H2SO4 solution will be neutralized.