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The reaction of sulfuric acid (H2SO4) with potassium hydroxide (KOH) is described by the equation below. Suppose
0.06 L of KOH with unknown concentration is placed in a flask with bromthymol blue indicator. A solution of 0.20 M
H2504 is dripped into the KOH solution. After exactly 0.017 L of H2SO4 is added, the Indicator changes from blue to
yellow. What is the concentration of the KOH? You must show all of your work to earn credit. (4 points)
H2SO4 + 2KOH → K2SO4 + 2H20

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Edufirst

Answer:

         [tex]\large\boxed{\large\boxed{1M}}[/tex]

Explanation:

1. Reaction

  [tex]H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O[/tex]

2. Mole ratio

    [tex]\dfrac{2molKOH}{1molH_2SO_4}[/tex]

3. Sulfuric acid

     [tex]Molarity=\dfrac{moles}{volume(liters)}[/tex]

     [tex]Moles=Molarity\times Volume(liters)[/tex]

     [tex]Moles=0.20M\times 0.017liter=0.034molH_2SO_4[/tex]

4. Potassium hydroxide

    [tex]\dfrac{2molKOH}{1molH_2SO_4}\times 0.034molH_2SO_4=0.068molKOH[/tex]

    [tex]Molarity=\dfrac{0.068mol}{0.06liter}=1.13M\approx 1M[/tex]

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