Question

What is the pH of the solution when 0.2M HCl, 0.4M NaOH and 0.2 M HCN is mixed, assuming the volume is constant. (Ka(HCN)= 5x10^-10)

Answer 1

for mixture Hol solution is Basic more and NaOH Resultant because the conc than strong and of Base is HCl + Naoh. Nac! + H₂O intial 0.2 0.um final 0 (0.4-02) 0.2 - 0.2 Remaing cone of [HO OH] = 0.214

Now for Nauh & HON HCN & NaOH H₂O Nach & intial 0.8 0.2 0 0 final o 0 0.2 0.2 m Since the at tinal Conc. of conc. of HCN & Nauh is zero state so, CN- & conc. of Nach [EN ] – 0.2 m CN- + H₂O HON & OH- 0 0 + n 0.2 - n Kb [HCN] [04] [CN-] ain 0-2-2 ( <su no

- 10 1 ke - Sя 10 ko'k. - lo-19 - 2 А 10 S10-10 я по- - 0,6 - 4-чт 0 3 Іоло 15 Sя 10-10 Къ - ,x 10s Сон:] - ч. 4+ 10- Рон - - lоо (ч.члів ) Рон = 2.35) PH + fон - фото 14 PH - Tч - Рон - ч - 2-3 PH - 11 63 Ang