Question

40 mL of a mixture of CH\(_3\)COOH and HCl (aqueous solution) is titrated against 0.1 M NaOH solution conductometrically. Which of the following statement is correct?

• A. The concentration of CH\(_3\)COOH in the original mixture is 0.005 M
• B. The concentration of HCl in the original mixture is 0.005 M
• C. CH\(_3\)COOH is neutralised first followed by neutralisation of HCl
• D. Point 'C' indicates the complete neutralisation HCl

Hint:

In the conductometric titration of a mixture of a strong acid and a weak acid with a strong base, the strong acid is neutralized first, leading to a decrease in conductance due to the replacement of highly mobile H\(^+\) ions. The weak acid is neutralized subsequently, leading to an increase in conductance due to the formation of a salt. The equivalence points can be identified from the changes in the slope of the conductance curve.

Answer 1

The Correct Option is B

The conductometric titration curve shows the change in conductance as NaOH solution is added to the mixture of HCl (strong acid) and CH\(_3\)COOH (weak acid). 

Region A-B: When NaOH is added to the mixture, it first neutralizes the strong acid, HCl, because the reaction between a strong acid and a strong base is more favorable. \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] During this neutralization, highly mobile H\(^+\) ions are replaced by less mobile Na\(^+\) ions, leading to a decrease in conductance. This corresponds to the region A-B of the curve. The volume of NaOH used to neutralize HCl is 2.0 mL (from the x-axis). 

Region B-C: After all HCl is neutralized, the added NaOH starts neutralizing the weak acid, CH\(_3\)COOH: \[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \] During this neutralization, weakly conducting CH\(_3\)COOH is converted into strongly conducting CH\(_3\)COONa (Na\(^+\) and CH\(_3\)COO\(^-\) ions). This leads to a gradual increase in conductance, as shown in the region B-C of the curve. The volume of NaOH used to neutralize CH\(_3\)COOH is (5.0 - 2.0) mL = 3.0 mL. 

Region C-D: After the complete neutralization of both HCl and CH\(_3\)COOH, further addition of NaOH (a strong electrolyte) leads to a sharp increase in conductance due to the increase in the concentration of highly mobile OH\(^-\) and Na\(^+\) ions. Now let's calculate the concentrations of HCl and CH\(_3\)COOH in the original mixture. For HCl: Moles of NaOH used = Molarity × Volume (in L) = 0.1 M × (2.0 / 1000) L = 0.0002 moles Since HCl and NaOH react in a 1:1 molar ratio, moles of HCl in the original 40 mL mixture = 0.0002 moles Concentration of HCl = Moles / Volume (in L) = 0.0002 moles / (40 / 1000) L = 0.0002 / 0.04 M = 0.005 M For CH\(_3\)COOH: Moles of NaOH used = Molarity × Volume (in L) = 0.1 M × (3.0 / 1000) L = 0.0003 moles Since CH\(_3\)COOH and NaOH react in a 1:1 molar ratio, moles of CH\(_3\)COOH in the original 40 mL mixture = 0.0003 moles Concentration of CH\(_3\)COOH = Moles / Volume (in L) = 0.0003 moles / (40 / 1000) L = 0.0003 / 0.04 M = 0.0075 M Based on these calculations: (1) The concentration of CH\(_3\)COOH in the original mixture is 0.0075 M, not 0.005 M. 

So, statement (1) is incorrect. (2) The concentration of HCl in the original mixture is 0.005 M. So, statement (2) is correct. (3) HCl, being a strong acid, is neutralized first because it reacts more readily with the strong base NaOH than the weak acid CH\(_3\)COOH. 

So, statement (3) is incorrect. (4) Point 'B' indicates the complete neutralization of HCl, and point 'C' indicates the complete neutralization of CH\(_3\)COOH. So, statement (4) is incorrect.