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.

Explanation:

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

And, HF and phenol are weaker acids having values of and respectively.

In the same way, methyl amine and pyridine are weaker bases with values of and respectively.

(i)   Volume will be calculated as follows.

Therefore, volume of NaOH is 100 ml.

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

(iii)  For Methyl amine,

                     =

And as,

Hence, the volume of HCl is 6.6 ml.

(iv) For HF,

                      =

As,    

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).

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