Question

Consider the following four titrations.
i. 100.0 mL of 0.10 M HCl titrated by 0.10 M NaOH

ii. 100.0 mL of 0.10 M NaOH titrated by 0.10 M HCl iii. 100.0 mL of 0.10 M CH3NH2 titrated by 0.10 M HCl

iv. 100.0 mL of 0.10 M HF titrated by 0.10 M NaOH Rank the titrations in order of:

increasing volume of titrant added to reach the equiva-

lence point.

increasing pH initially before any titrant has been added.

increasing pH at the halfway point in equivalence.

increasing pH at the equivalence point.

How would the rankings change if C5H5N replaced CH3NH2 and if HOC6H5 replaced HF?

Answer 1

(a)

HCl is a stronger acid.

NaOH is a stronger base.

HF and Phenol are weaker acids with K_a values of 6.6 x 10^-4 and 1.3x10^-10respectively.

similarly, Methyl amine and pyridine are weaker bases with K_b values of 4.4x10^-4 and 1.7x10^-9 respectively.

(i) M_aV_a = M_b V_b

100 x 0.1 =0.1 x V

So Volume of NaOH = 100ml.

(ii) Similarly, volume ofHCl = 100ml.

(iii) For Methyl amine, [OH]^- =4.4x10^-4 x 0.1 =6.6 x 10^-3

M_a V_a = M_b V_b

0.1 x V = 6.6 x 10^-3 x 100

Volume of HCl = 6.6 ml

(iv) For HF , [H]⁺ =  6.6 x 10^-4 x0.1 = 8.12 x 10^-3

M_a V_a = M_b V_b

8.12x10^-3 x 100 = 0.1 x V

Volume of NaOH added = 8.12ml.

(b) the increasing order of volumsof titrant is (i) = (ii) > (iv) >(iii)

(c) p^H values of (i) < (iv) < (iii) <(ii), because p^H  of Stronger base > weaker base > weaker acid > Stronger acid

(d) at equivalence point p^H   of all the solutions become equal.

(e)When Pyridine and Phenol are used, volumes of titrants decrease as their K_a and K_b values are lesser than methyl amine and phenol