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

Step-by-step 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 * 10^(-4)` and
`1.3 * 10^(-10)` respectively.

In the same way, methyl amine and pyridine are weaker bases with
`K_(b)` values of
`4.4 * 10^(-4)` and
`1.7 * 10^(-9)` respectively.

(i) Volume will be calculated as follows.

`M_(a)V_(a) = M_(b)V_(b )`

`100 * 0.1 = 0.1 * V`

Therefore, volume of NaOH is 100 ml.

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

(iii) For Methyl amine,

`[OH]^(-) = \sqrt{4.4 * 10^(-4) * 0.1}`

=
`6.6 * 10^(-3)`

And as,
`M_(a)V_(a) = M_(b)V_(b)`

`0.1 * V = 6.6 * 10^(-3) * 100`

Hence, the volume of HCl is 6.6 ml.

(iv) For HF,

`[H]^(+) = \sqrt{6.6 * 10^(-4) * 0.1}`

=
`8.12 * 10^(-3)`

As,
`M_(a)V_(a) = M_(b)V_(b)`

`8.12 * 10^(-3) * 100 = 0.1 * 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).