Question

consider the titration of 500mL of HCl with the pH of 3.52. How much 0.04M NaOH must be added to reach the equivalence point.

Answer 1

HCl is a strong acid . Consider dissociation of HCl in water.

HCl ( aq) + H₂O (l) H₃O ⁺ (aq) + Cl ^- (aq)

From above reaction, [ HCl ] = [ H₃O ⁺ ]

We can calculate [ H₃O ⁺ ] from pH of solution.

We have relation, pH = - log [ H₃O ⁺ ]

[ H₃O ⁺ ] = 10 ^-pH

[ H₃O ⁺ ] = 10 ^{- 3.52} = 3.02 10 ^-04 M

We have, [ HCl ] = [ H₃O ⁺ ] , therefore [ HCl ] is 3.02 10 ^-04 M .

Consider a reaction, HCl + NaOH NaCl + H₂O

From reaction, stoichiometric ratio = No of moles of acid / No of moles of base

stoichiometric ratio = No. of moles of HCl / No. of moles of NaOH = 1/1=1

We have correlation, M _acid V _acid = M _base V _base stoichiometric ratio.

Therefore, 3.02 10 ^-04 M 500 ml = 0.04 M V _base 1

V _base = 3.02 10 ^-04 M 500 ml / 0.04 M 1

V _base =3.775 ml

ANSWER : Volume of NaOH required to reach the equivalence point is 3.8 ml