Question

(a) Calculate pH of 1.0 × 10^–8 M solution of HCl.

(b) The species: H₂O, HSO₄^– and NH₃ can act both as Bronsted acids and bases. For each case, give the corresponding conjugate acid and conjugate base.

Answer 1

(a) H₂O(l) + H₂O(l) ⇌ H₂O⁺ (aq) + OH^– (aq)

K_w = [H₃O⁺ ] [OH^–]

= 10^–14

Let x = [OH^–] = [H₃O⁺ ] from H₂O

The H₃O⁺ concentration is generated (i) from the ionization of HCl dissolved i.e.

HCl(aq) + H₂O(l) ⇌ H₃O + (aq) + Cl^– (aq) and (ii) from ionization of H₂O.

In these very dilute solutions, both sources of H₃O⁺ must be considered:

[H₃O⁺] = 10^–8 + x

\(\therefore\) K_w = (10^–8 + x) (x) = 10^–14

or x² + 10^–8 x – 10^–14 = 0

The value of x can be determined from quadratic equation

x = 9.5 × 10^–8

[OH^– ] = 9.5 × 10^–8

Now, pOH = – log [OH^– ]

= – log 9.5 + 8 log 10

= – 0.098 + 8

pOH = 7.02

\(\because\) pH + pOH = 14

pH + 7.02 = 14

or 

pH = 14 – 7.02 = 6.98

Hence, pH of 10–8 M solution of HCl is 6.98.

(b)

Species	Conjugate acid	Conjugate base
H2O HSO4– NH3	H3O+ H2SO4 NH4+	OH– SO42– NH2–