40 mL of a mixture of CH3COOH and HCl (aqueous solution) is titrated against 0.1 M NaOH solution conductometrically. Which of the following statement is correct? [figure]
- The concentration of CH3COOH in the original mixture is 0.005 M
- The concentration of HCl in the original mixture is 0.005 M
- CH3COOH is neutralised first followed by neutralisation of HCl
- Point 'C' indicates the complete neutralisation of HCl
Answer
(2) The concentration of HCl in the original mixture is 0.005 M
Solution
A → B: HCl (strong acid) is neutralised first; highly mobile H+ ions are replaced by slower Na+ ions conductance falls.
B (2.0 mL): HCl completely neutralised.
B → C: CH3COOH is neutralised; ionic CH3COO- Na+ is formed conductance rises slowly.
C (5.0 mL): CH3COOH completely neutralised; beyond C excess OH- sharp rise (C → D).
HCl: M 40 = 0.1 2.0 MHCl = 0.005 M (2) is correct.
CH3COOH: M 40 = 0.1 (5.0 - 2.0) M = 0.0075 M (1) is incorrect.
(3) is incorrect (HCl is neutralised first) and (4) is incorrect (point C is the end point of CH3COOH; HCl ends at B).
Answer: (2)