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Jenna
Jenna asked in Science & MathematicsChemistry · 6 years ago

A volume of 90.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4).?

A volume of 90.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 22.2 mL of 1.50 M H2SO4 was needed? The equation is

2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

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  • Sheldon
    Lv 5
    6 years ago
    Favorite Answer

    2 KOH + H2SO4 → K2SO4 + 2H2O

    moles H2SO4: 0.0222 L x 1.50 M= 0.0333

    from bal. rxn., 1 mole H2SO4 reacts with 2 moles KOH, 1:2

    0.0333 moles H2SO4 x 1/2= 0.01665 moles KOH

    0.01665 moles KOH= 0.090 L x M

    M KOH= 0.185

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