Question

Consider the following four titrations: i. 100.0 mL of 0.10 M HCl titrated with 0.10 M NaOH ii. 100.0 mL of 0.10 M NaOH titrated with 0.10 M HCl iii. 100.0 mL of 0.10 M CH3NH2 titrated with 0.10 M HCl iv. 100.0 mL of 0.10 M HF titrated with 0.10 M NaOH Rank the titrations in order of increasing volume of titrant added to reach the equivalence point.

Answer 1

Explanation:

As we know that HCl is a stronger acid  and NaOH is a stronger base.

And, HF and phenol are weaker acids having $K_{a}$ values of $6.6 \times 10^{-4}$ and $1.3 \times 10^{-10}$ respectively.

In the same way, methyl amine and pyridine are weaker bases with $K_{b}$ values of $4.4 \times 10^{-4}$ and $1.7 \times 10^{-9}$ respectively.

(i)   Volume will be calculated as follows.

           $M_{a}V_{a} = M_{b}V_{b }$

           $100 \times 0.1 = 0.1 \times V$

Therefore, volume of NaOH is 100 ml.

(ii)   Similarly, volume of HCl is 100 ml.

(iii)  For Methyl amine,

         $[OH]^{-} = \sqrt{4.4 \times 10^{-4} \times 0.1}$

                     = $6.6 \times 10^{-3}$

And as, $M_{a}V_{a} = M_{b}V_{b}$

             $0.1 \times V = 6.6 \times 10^{-3} \times 100$

Hence, the volume of HCl is 6.6 ml.

(iv) For HF,

           $[H]^{+} = \sqrt{6.6 \times 10^{-4} \times 0.1}$

                      = $8.12 \times 10^{-3}$

As,    $M_{a}V_{a} = M_{b}V_{b}$

         $8.12 \times 10^{-3} \times 100 = 0.1 \times V$

Hence, the volume of NaOH added is 8.12 ml.

Therefore, we can conclude that the increasing order of volums of given titrant is (i) = (ii) > (iv) > (iii).