A volume of 80.0 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 13.7 mL of 1.50 M H2SO4 was needed? The equation is

The equation is H2SO4 + 2KOH   ===>   K2SO4 + 2H2O

The molarity of base is 0.257 M

Explanation:

Using titration equation CAVA/CBVB = NA/NB

Where NA is the number ov mole of acid =1

Using titration equation

CAVA/CBVB = NA/NB

Where NA is the number of mole of acid = 1

NB is the number of mole of base = 2

CA is the molarity of acid =1.5M

CB is the molarity of base = to be calculated

VA is the volume of acid = 13.7 mL

VB is the volume of base = 80 mL

Substituting

1.5×80/CB×13.7 = 1/2

Therefore CB = 1.5×13.7×2/80×1

CB = 0.257 M.

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.