acid base equilibrium @ nzphddr的部落格

標題:

acid base equilibrium

發問:

Calculate the pH of a 0.01M solution of acid HA (given Ka of HA =1.7*10^-10)Both[H+] and [OH-] can be calculatedH+ is come from the dissociation of HA(HA=A- + H+)but where is [OH-]come from?? and we can still calculate [OH-]because I can't see any [OH-]given out at the dissociation, only [H+] is given... 顯示更多 Calculate the pH of a 0.01M solution of acid HA (given Ka of HA =1.7*10^-10) Both[H+] and [OH-] can be calculated H+ is come from the dissociation of HA(HA=A- + H+) but where is [OH-]come from?? and we can still calculate [OH-] because I can't see any [OH-]given out at the dissociation, only [H+] is given out.

最佳解答:

In the solution, water undergoes self-ionization to give H+(aq) and OH- ions. H2O(l) ≒ H+(aq) + OH-(aq) Kw = [H+][OH-] = 1 x 10-14 M2 Since Kw is too small, the concentration of H+(aq) ions depends almost entirely on the dissociation of HA(aq). However, OH-(aq) ions also exist in the solution due to the self-ionization of water. ======== Consider the dissociation of HA: startaa aaaaHA(aq) ≒ A-(aq) + H+(aq) start aaaa 0.01 Mq) + HA0 Mq) + H0 M change aaa -y M a(l) ≒ +y M) +H+y M eqm aaaa(0.01 - y) M+HAy M) + H3y M Ka = y2 / (0.01 - y) = 1.7 x 10-10 (M) y2 + (1.7 x 10-10)y - (1.7 x 10-12) = 0 [H+] = y = 1.3 x 10-6 M Consider the self-ionization of water: eqm H2O(l) ≒ OH-(aq) + H+(aq) eqm 2H2O(l)≒ H? M q 1.3 x 10-6 M Kw = [OH-][H+] [OH-] = Kw/[H+] = (1 x 10-14)/(1.3 x 10-6) = 7.69 x 10-9 M

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