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.

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b