calculate the pH of a 1.94M solution of nitrous acid (HNO2 - ka = 4.5E-4).
Set up an ICE chart, something like the following:
............HNO2 ==> H^+ + NO2^-
initial....1.94M.....0.......0
change......-x.......+x.......+x
final......1.94-x.....x........x
Ka = (H^+)(NO2^-)/(HNO2)
Substitute from the ICE chart to the Ka expression, solve for x and convert to pH.
for .148M it is 2.00
To calculate the pH of a solution of nitrous acid (HNO2), we need to use the equation for the ionization of the acid and the acid dissociation constant (Ka).
The ionization reaction for nitrous acid is:
HNO2 ⇌ H+ + NO2-
The Ka expression for this reaction is:
Ka = [H+][NO2-] / [HNO2]
Given that Ka = 4.5E-4 and the initial concentration of nitrous acid is 1.94 M, we can assume that the concentration of H+ and NO2- are both x M at equilibrium, and the concentration of HNO2 will be (1.94 - x) M.
Now, substitute these values into the Ka expression:
4.5E-4 = x * x / (1.94 - x)
Since x is much smaller than 1.94, we can assume that (1.94 - x) ≈ 1.94:
4.5E-4 = x^2 / 1.94
Rearranging the equation gives us:
x^2 = 4.5E-4 * 1.94
x^2 = 8.73E-4
x = √(8.73E-4)
x ≈ 0.029 M
So, the concentration of H+ is approximately 0.029 M in the solution.
To find the pH, we can use the formula:
pH = -log10[H+]
Substituting the value we found:
pH = -log10(0.029)
pH ≈ 1.54
Therefore, the pH of a 1.94 M solution of nitrous acid is approximately 1.54.