Question

I. REACTION EQUILIBRIUM IN NON-IDEAL SYSTEMS The acetic acid (CH:COOH) aqueous solution with a concentration of 0.1 mol/kg has a hydrogen ion concentration of 1.35x10-3 mol/kg. i. Write the acid dissociation equation. ii. Calculate the acid dissociation constant, Ka, considering the ionic activity. jïi. Calculate the pH of the acidic solution. iv. What is the new pH if the solution contains 0.1 mol CH3COOH and 0.4 mol of sodium acetate, CH3COONa, in 1.0 L of water? Consider the negligible quantitities. V. Explain briefly why the pH of the two solutions in iii) and iv) are different from each other?

Answer 1

at = the I-(i) CH3COOH CH3COO70q) + Hteera) (1) Ka= [H + Jeg. [CH₃Cooler [CH₃CO Other where [Jeg means concentration equilibrium So [H+Jegast = 135x 10-3 mol/ kg [CH3Coo-Jeg CH+Jeg= 1.38x10-3mad kg [CH3COOH] wol mol col mol/kg So, ka = (1.35x10-3² mal/kg 1.823x10-5 mallkg pH of the acidic sol" at equilibrium is a pH= -dog10 (1:35x10-3 2,8 (iv) pH of the solh containing and mol CH₂COOH and a. 4 mal of CH3COONa in IL of pa pH of buffer is given as: pH = pkalche footh + logro [CH3COONa] [CH3 Cool -Loge CH + water is:

pH = -log Llog (1.823x10-5) + legio = 4.739 + 0.602 = 5.34) (0.4M) (6.IM) امیلیا به us - The pH of the two solis in ii) and (iv) are different from each other at is bec ause in acid, Ht gets dissociated in water and giver the pH. Whereas, in the buffer, et it dissociated from the acid is partially consumed by the conjugate base Chere, CH₃COONa) pH of liv) is different/ desser than (iii). and so