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The reaction of sulfuric acid (H2SO4) with potassium hydroxide (KOH) is described by the equation:

H2SO4 + 2KOH → K2SO4 + 2H2O

Suppose 50 mL of KOH with unknown concentration is placed in a flask with bromthymol blue indicator. A solution of 0.10 M H2SO4 is dripped into the KOH solution. After exactly 33.4 mL of H2SO4 is added, the indicator changes from blue to yellow. What is the concentration of the KOH?

A. 0.067 M
B. 0.13 M
C. 0.15 M
D. 0.26 M

Respuesta :

Answer:

B. 0.13 M

Explanation: Because 1 mole of sulfuric acid reacts with 2 moles of potassium hydroxide, there are twice as many moles of KOH at the equivalence point as moles of H2SO4.                  

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