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6 January, 01:12

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?

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  1. 6 January, 01:17
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    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.
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