Answer
3.35
Steps
Open sidebar
NO2 pH Calculation.
what is the pH of a 0.25 M solution of NO2 (Ka=4.5x10^-4)
NO2 is a weak acid, and its dissociation in water can be represented by the following chemical equation:
NO2 + H2O ⇌ HNO2 + OH-
The equilibrium constant for this reaction is the acid dissociation constant (Ka) for HNO2, which is given as 4.5 x 10^-4.
To calculate the pH of a 0.25 M solution of NO2, we need to find the concentration of H+ ions in the solution. We can do this by using the equilibrium constant expression for the reaction:
Ka = [HNO2][OH-]/[NO2]
Assuming that the concentration of H+ ions from water dissociation is negligible compared to the concentration of OH- ions from the dissociation of NO2, we can simplify the expression as follows:
Ka = [HNO2][OH-]/[NO2] ≈ [HNO2][OH-]/[HA]
where HA represents the concentration of undissociated NO2.
Since NO2 is a weak acid, we can assume that the concentration of undissociated NO2 is approximately equal to the initial concentration of NO2, which is 0.25 M.
Therefore, we can rearrange the equilibrium constant expression and solve for [H+]:
Ka = [HNO2][OH-]/[NO2]
[HNO2] = Ka*[NO2]/[OH-]
[H+] = Ka*[NO2]/[OH-]
Now we need to find the concentration of OH- ions in the solution. Since the reaction between NO2 and water produces OH- ions, we can use the conservation of mass principle to calculate their concentration.
Initially, the concentration of OH- ions is zero, so we can assume that all of the OH- ions produced come from the dissociation of NO2. Therefore, we can write:
[OH-] = [HNO2]
Substituting this expression into the equation for [H+], we get:
[H+] = Ka*[NO2]/[HNO2]
[H+] = Ka*[NO2]/([NO2] + [HNO2])
[H+] = 4.5 x 10^-4 * 0.25 / (0.25 + x)
where x is the concentration of HNO2 that has dissociated.
Since the dissociation of NO2 is small compared to its initial concentration, we can assume that the concentration of HNO2 that has dissociated is small compared to 0.25 M. Therefore, we can approximate [HNO2] + [NO2] ≈ 0.25 M.
[H+] = 4.5 x 10^-4 * 0.25 / 0.25
[H+] = 4.5 x 10^-4
Taking the negative logarithm of both sides of the equation, we get:
pH = -log[H+]
pH = -log(4.5 x 10^-4)
pH = 3.35
Therefore, the pH of a 0.25 M solution of NO2 is approximately 3.35.
ChatGPT