Question

We place 0.577 mol of a weak acid, HA, and 13.6 g of NaOH in enough water to produce 1.00 L of solution. The final pH of this solution is 4.50 . Calculate the ionization constant, K_a, of HA.

Answer 1

Mole of HA=0.577 mole Mass of NaOH= Mole of NaOH = = mole of [our . Naot is strong electrolyle The Neutralization Reaction is : Htt OH - thot 10+ H+(excss) 13.6 gm 18.6 gm 40 gmsomol 0.34 mole 80. 80 H+.(excess can be Calculated by PH) PH= -log[et] » -PH [ht] = 10 104.5 [H]_excess) [H+] = 3.168105 mde/L 80 Total [H+] = 3.16810S + Reacted with Cour] Total [++] = [3.16X10S + 0.34] [ht] = 0.340032 M 0 since HA - Ht & A Inital 0.577M ateq: (0.577-x) and x= 0.340032 M Ka= (0.340032) 2 (0.577–0-340032) 2) Ika= 0.488 80