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(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. |
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Answer» (a) H2O(l) + H2O(l) ⇌ H2O+ (aq) + OH– (aq) Kw = [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 \(\therefore\) Kw = (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)
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