Calculate the % dissociation given 1.4M HOCl (Ka=3.5x10^-8) .02M HOCl 2.1M HF (K

Question

Calculate the % dissociation given 1.4M HOCl (Ka=3.5x10^-8) .02M HOCl 2.1M HF (Ka=3.5x10^-4) Please do it step by step! If you do, feel free to only answer one and I can do the rest! I'm just not sure how to approach problem! Thanks and God Bless You!!

Explanation / Answer

HOCl---> H+ OCl- , [HOCl= 1.4-x , [H+]=[OCl-] = x , Ka = [H+][OCl-]/[HOCl] = ( x^2)/(1.4-x) = 3.5 x10^ -8 , x^2 +(3.5 x10^ -8x) -(4.9 x10^ -8) = 0 , x = 0.00022 = [H+] , % dissociation = 100 x [H+]/[HOCl] = 100x0.00022/(1.4-0.00022) = 0.0157 % , similarly others can be found out.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.