Answer:
When we add 0.33L of KOH only the HCl solution will be neutralized. When we add <u>1.4 L</u> of KOH, all of the acid (the HCl and H2SO4 solution) will be neutralized.
Explanation:
<u>Step 1:</u> 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
<u>Step 2: </u>Calculate moles of HCl
Moles HCl = Molarity HCl * volume
Moles HCl = 0.100 M * 0.5 L
Moles HCl = 0.05 moles
<u>Step 3:</u> Calculate moles of H2SO4
Moles H2SO4 = 0.210 M * 0.5 L
Moles H2SO4 = 0.105 moles
<u>Step 4:</u> 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
<u>Step 5:</u> 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
<u>Step 6: </u>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.
