Question

2. The pk, of hypochlorous acid (HOCI) is 7.53. (a) Calculate the k. of HOCI. Given that the pk, of hypobromous acid (HOBr) is 8.63, which acid is stronger, and how do you know? (b) Write the reaction (with physical states) that corresponds to the K, of HOCI. Identify the weak acid among the reactants (label it WA) and its conjugate base among the products (label it CB). (c) Calculate [H30*] (or "[H]") and pH for a 0.15 M HOCI solution. Check that any approximations you make are valid (cause no more than 5% error). (d) What percent of the total HOCI has dissociated in a 0.15 M HOCI solution? Would the percent dissociation increase or decrease upon dilution, and why?

Help please and show all work!

Answer 1

2. (a) Given, the pKa of hypochlorous acid HOCl = 7.53 and the pKa of hypobromous acid HOBr = 8.63

For HOBr	For HOCl
pKa = 8.63 or, - log Ka = 8.63 or, Ka = 10 - 8.63 ∴ Ka = 2.3×10-9	pKa = 7.53 or, - log Ka = 7.53 or, Ka = 10 - 7.53 ∴ Ka = 2.95×10-8

The required value of K_a of HOCl = 2.95×10^-8

We know that, greater value of K_a (i.e. lower value of the pK_a), stronger the acid due to greater the ability to donate H⁺ ions in aqueous solution.

Since, pK_a (HOCl) < pK_a (HOBr)

So, HOCl is stronger acid than that of HOBr.

(b) The reaction related to K_a is given by-

HOCl (aq) +	H2O (l)	⇌	H3O+ +	OCl-
(WA)				(CB)

                                                K_a = [H₃O⁺] [OCl^-] / [HOCl] [H₂O]

(c) [H₃O⁺] or [H⁺] can be calculated as follows:

            [H⁺] = √(K_a × C_a)       

where,             K_a = dissociation constant = 2.95×10^-8

and C_a = conc. of acid            = 0.15 M

or, [H⁺] = √(2.95×10^-8 × 0.15)

or, [H⁺] = √(4.4×10^-9) = 6.6×10^-5

∴ [H₃O⁺] or [H⁺] = 6.6×10^-5

∴ pH = -log [H⁺] = - log [6.6×10^-5] = 4.18

(d) Degree of dissociation (α) = [H⁺] / C_a = (6.6×10^-5) / (0.15) = 0.00044

Percent of dissociation = (0.00044/0.15) × 100 % = 0.29 %

With dilution % of dissociation will increase.

Because with dilution conc. of water increases and equilibrium shift towards right side. Thus, increases the donation of proton and hence % of dissociation increases.