A volume of 40.0 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 16.2 mLmL of 1.50 MM H2SO4H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l )

The concentration of KOH solution is 1.215 M

Explanation:

For the given chemical equation:

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:

Hence, the concentration of KOH solution is 1.215 M

i believe it is 40–40–20%

explanation: