Calculate [H3O+]
0.100 M NaNO2 and 5.00×10−2 M HNO2
so there's no Ka value for NaNO2, it can be ignored
5.6x10^-4 = x^2 / 5.00x10^-2
x = 5.3x10^-3
the answer is incorrect why? Thanks
A mixture of a weak acide (HNO2) and its salt (NaNO2) is a buffer solution and you should be using the Henderson-Hasselbalch equation.
pH = pKa + log (base)/(acid). NaNO2 is the base and HNO2 is the acid.
Find pH and convert to H3O^+.
To calculate the concentration of H3O+ in a solution, you need to consider the ionization of the weak acid, HNO2. The equation for the ionization of HNO2 is as follows:
HNO2 + H2O ⇌ H3O+ + NO2-
The initial concentration of HNO2 is given as 5.00×10−2 M. Let's assume the concentration of H3O+ formed is x.
Since HNO2 is a weak acid, we can use the acid dissociation constant (Ka) expression to calculate the concentration of H3O+. The Ka expression for HNO2 is as follows:
Ka = [H3O+][NO2-] / [HNO2]
However, note that NO2- does not affect the concentration of H3O+ in this case, so we can ignore it. The equation simplifies to:
Ka = [H3O+][NO2-] / [HNO2]
Ka = [H3O+] / [HNO2]
We are given that the Ka value for HNO2 is 5.6×10^-4. Substituting this into the equation:
5.6×10^-4 = [H3O+] / [HNO2]
5.6×10^-4 = x / 5.00×10^-2
Now, we can solve for x:
x = 5.6×10^-4 * 5.00×10^-2
x ≈ 2.8×10^-6
So, the correct concentration of H3O+ in the solution is approximately 2.8×10^-6 M.