Question

Find the pH of a solution resulting when 50.0 mL of 0.20M HCl is mixed with 50.0 mL of 0.20M HC₂H₃O₂. K_a=1.8x10^-5

Answer 1

CH3COOH---------->CH3COO- + H+, HCl suppliments H+ions.

moles of CH3COOH before mixing = Moarity* Volume(L)= 0.2*50/1000 =0.01 moles

Moles of HCl = 0.2*50/1000 =0.01 moles

Volume after mixing = 50+50= 100ml =0.1L

after mixing : Concentration of CH3COOH= 0.01/0.1= 0.1M and HCl=0.1M

let x= drop in concentration of CH3COOH to reach equilibrium

so at equilibrium [H+] =0.1+x and CH3COO- =x

hence Ka= 1.8*10^-5= [CH3COO-] [H+]/[CH3COOH] = (0.1+x)*x/(0.1-x), when solved using excel, x= 1.8*10^-5

pH= -log (1.8*10-5) =4,7447