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

Question

Chemistry

Meredith Pierce

4 May, 03:05

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

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Noel Ponce · Answer 1

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.