What is the pH of 0.10 M HNO2 given Ka = 4.5x10-4.
I got 2.17 as my final answer. Is that correct?
HNO2 ==> H^+ + NO2^-
Ka = (H^+)(NO2^-)/(HNO2)
Set up an ICE chart and substitute into Ka. Solve for H^+, then convert to pH.
Yes that's correct
To find the pH of a solution of a weak acid like HNO2, you need to use the equilibrium expression for the acid dissociation reaction (Ka) and solve for the concentration of H+ ions.
The equilibrium expression for the dissociation of HNO2 is:
HNO2 ⇌ H+ + NO2-
The Ka expression for HNO2 is:
Ka = [H+][NO2-]/[HNO2]
Given that the concentration of HNO2 is 0.10 M and Ka = 4.5x10^-4, we can assume that x is the concentration of H+.
Therefore, at equilibrium, [H+] = x and [NO2-] = x.
Since the initial concentration of HNO2 is equal to the initial concentration of H+, we can substitute [HNO2] with 0.10 - x.
Now, the equilibrium expression becomes:
Ka = (x)(x) / (0.10 - x)
Simplifying the equation:
Ka(0.10 - x) = x^2
Now, let's make an assumption based on the given information. Ka is small compared to the initial concentration of HNO2, so we can assume that x is small compared to 0.10 M. This means that we can neglect x in comparison to 0.10 M when subtracting from 0.10 M. Therefore, we can assume that 0.10 - x is approximately equal to 0.10.
Now, the equation simplifies to:
Ka(0.10) ≈ x^2
0.10 * Ka ≈ x^2
Taking the square root of both sides:
√(0.10 * Ka) ≈ x
Now, plug in the value of Ka = 4.5x10^-4 and solve for x:
√(0.10 * 4.5x10^-4) ≈ x
x ≈ 0.0212
Therefore, the concentration of H+ ions, and thus the pH, is approximately 0.0212 M.
To calculate the pH, you can take the negative logarithm (base 10) of the H+ concentration:
pH = -log10(0.0212) ≈ 1.67
So, the pH of the 0.10 M HNO2 solution is approximately 1.67.