A volume of 90.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2SO4). What was the molarity of the KOHKOH solution if 24.7 mLmL of 1.50 MM H2SO4H2SO4 was needed

0.823 M was the molarity of the KOH solution.

Explanation:

(Neutralization reaction)

To calculate the concentration of base , we use the equation given by neutralization reaction:

where,

are the n-factor, molarity and volume of acid which is

are the n-factor, molarity and volume of base which is KOH.

We are given:

Putting values in above equation, we get:

0.823 M was the molarity of the KOH solution.

pie chart because it is in percents ( % )

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