Aquality control chemist at dow chemical tried to determine purity of naoh using titration. he measured out 0.500 g naoh sample and dissolved it in 20 ml water. 22.5 ml of a 0.500 mol/l hcl solution was used to reach the end point. assume the impurity did not react with hcl, what is the % purity of this naoh sample?

89.94 %

Explanation:

Considering:

Or,

Given :

For HCl :

Molarity = 0.500 M

Volume = 22.5 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 22.5×10⁻³ L

Thus, moles of HCl :

Moles of HCl = 0.01125 moles

The reaction of NaOH and HCl is:

NaOH + HCl ⇒ NaCl + H₂O

Moles of HCl = Moles of NaOH

Thus, Moles of NaOH = 0.01125 moles

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

Thus,

Total mass = 0.500 g

% purity = ( 0.4497 / 0.500 ) × 100 = 89.94 %

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.