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 %

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.