Question

Obliczyc stopień dysocjacji HCN (K= 7,5* 10^-10) w roztworze o pH = 5.

Answer 1

K = [H+][CN-]/[HCN]

pH = 5 ---> [H+] = 1*10(-5) mol/dm3

K = [H+]²/[HCN]
[HCN] = [H+]²/K = 1*10(-10)/7,5*10(-10) = 0,13 mol/dm3
C = [HCN]+ [H+] = 0,13

alfa = [H+]/C = 7,7*10(-5)

Proszę :)