The venom of biting ants contain formic acid, HCOOH (Ka = 1.80 * 10 ^-4) at 25 C. What is the pH of a 0.065 M solution of formic acid?
let the ionization equation be
HA ↔ H+ + A-
then
Ka = [H+] [A-] / [HA]
let [H+] = X, then [A-] = X and [HA] = (0.065 – X)
1.80E-4 = X^2 / (0.065 – X)
1.17E-5 – 1.80E-4 X = X^2
X^2 + 1.80E-4 X – 1.17E-5 = 0
X = 3.33E-3
pH = -log[H+] = -log 3.33E-3 = 2.48 (this has 2 sif figs)
well its a weak acid therefore [H+] = [COO-]
square brackets indicate concentration of by the way.
Ka = ( [H+][COO-] ) / [HCOOH] = [H+]^2 / [HCOOH]
rearrange to find [H+]^2 :
[H+]^2 = Ka * [HCOOH]
[H+] = ROOT{ Ka X*[HCOOH]
[H+] = ( (1.8*10^-4)*(0.065) )^0.5
[H+] = 1.17*10^-5
pH = -log[H+] = -log(1.17*10^-5) = 4.93 (3sf)
i used to hate this part of chem, but it gets really easy after a while when you nail it so don’t worry
…wat
