Question

I got x=0.011 and I know that -log(0.011)= the pH that is the answer but how do i know that i use 0.011 and why do i not divide 1.01x10^-14 by that like I do in different problems, or why do i not substitute the x back into the 0.5-x?

26) Farmers who raise cotton once used arsenic acid, H3 AsO4, as a defoliant at harvest time. Arsenic acid is a polyprotic acid with Kal = 2.5 x 10-4, Ka2 = 5.6 x 10-8, and Ka3 = 3 × 10-13. What is the pH of a 0.500 M solution of arsenic acid? - SocUE Kar ONLY A) 0.85 - HA ² TA x² B) 3.90 I 0.5 o o Lai= 0.50-X C) 4.51 . - x x x D) 1.95 É 0,5-8 E) None of these choices is correct. X² HOLDS X=0,010 M CAPPROK PH=1.95

Answer 1

Kw = [H+] [OH-]= 1.01 × 10^-14

pH = -log[H+]

We have arsenic acid it dissociates to give arsenate ion and H+. Ka1

While calculating the concentration of H+ ion generally the first dissociation constant is taken into account as the subsequent dissociation is very less.

By looking at the IEC table you can see that at equilibrium concentration of H+ is given as x.

So there is no need to divide by 1.01 × 10^-14 as that will give us the concentration of OH- ion.

[HA] = 0.5-x

We don't substitute the value of x in this as we just need the concentration of H+ ion which is given by x.