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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:

  • 1.13M

Explanation:

The change of color indicates that the solution was neutralized.

1. Neutralization equation(given):

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

2. Mole ratio:

  • 2moles of KOH are neutralized by 1 mol of H₂SO₄

3. Number of  moles of H₂SO₄

  • Molarity, M = 0.20M
  • Volume, V = 0.017 liter
  • M = n/V ⇒ n = M×V =0.20M × 0.017 liter = 0.034mol

4. Number of moles of KOH

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

5. Concnetration KOH

  • n = 0.0680 mol
  • V = 0.06 liter

  • M = n/V = 0.068 mol / 0.06 liter = 1.13 M ← answer

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