What is the hydronium ion concentration in a solution 0.150M in HNO2 and 0.300M in NaNO2? Ka = 7.24E-4
Use the Henderson-Hasselbalch equation.
Thanks!
To find the hydronium ion concentration in the solution, we first need to understand the dissociation of HNO2 in water.
The balanced equation for the dissociation of HNO2 in water is as follows:
HNO2 + H2O ⇌ H3O+ + NO2-
We need to use the equilibrium constant expression (Ka) to determine the concentration of hydronium ions (H3O+) in the solution. The expression for Ka is:
Ka = [H3O+][NO2-] / [HNO2]
Given that the Ka value is 7.24E-4, and the concentrations of HNO2 and NaNO2 in the solution are 0.150M and 0.300M, respectively, we can set up an ICE (Initial-Change-Equilibrium) table:
HNO2 + H2O ⇌ H3O+ + NO2-
Initial: 0.150M 0M
Change: -x +x
Equilibrium: 0.150M - x x x
Since the stoichiometric coefficient of HNO2 is 1, we can say that the initial concentration of HNO2 (0.150M) is equal to the equilibrium concentration (0.150M - x).
Substituting the equilibrium concentrations into the Ka expression:
Ka = (x)(x) / (0.150 - x)
Simplifying the equation:
7.24E-4 = x^2 / (0.150 - x)
Since the value of x is expected to be small compared to 0.150, we can assume that (0.150 - x) ≈ 0.150.
7.24E-4 = x^2 / 0.150
Rearranging the equation to find x:
x^2 = (7.24E-4)(0.150)
x^2 = 1.086E-4
x = sqrt(1.086E-4)
x ≈ 0.01042M
So, the concentration of H3O+ (hydronium ions) in the solution is approximately 0.01042M.