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.

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.