Avolume of 90.0 ml of aqueous potassium hydroxide (koh) was titrated against a standard solution of sulfuric acid (h2so4). what was the molarity of the koh solution if 25.2 ml of 1.50 m h2so4 was needed? the equation is 2koh(aq)+h2so4(aq)→k2so4(aq)+2h2o(l )

The molarity of KOH is 0.84 M.

Explanation:

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 molarity of KOH is 0.84 M.

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.