Buffers- Common Ion effect
IS this Correct
Calculate the pH of a aqueous solution containing 0.15 M HNO2 and 0.20 M NaNO2 (aq). The Ka of nitrous acid is 4.0*10^-4.
HNO2 + H20 <--> NaNO2- + H30^+
Inital 0.15 0.20 X
Final 0.15 0.20
Ka= [H30+][NaNO2]/ [HNO2]
4.0*10^-4 = [H30+][0.20]/[0.15]
3.0*10^ -4 = H30+
pH= 3.52
The answer is correct. The methodology is not.You made several errors (equation, Ka expression wrong, etc and other errors in substitution but all of the errors canceled out to give you the correct answer). Use the Henderson-Hasselbalch equation.
pH = pKa + log (base/acid)
pH = 3.40 + log(0.2/0.15) = 3.52
To calculate the pH of the given aqueous solution, we can use the Henderson-Hasselbalch equation. However, in this given case, we have a buffer solution consisting of a weak acid (HNO2) and its conjugate base (NaNO2), so we can use the common ion effect to simplify the equation.
The common ion effect occurs when a solution contains a weak acid and its conjugate base or a weak base and its conjugate acid. In this case, we have HNO2 as a weak acid and NaNO2 as its conjugate base.
To start, let's write the balanced equation for the dissociation of HNO2:
HNO2 + H2O ⇌ H3O+ + NO2-
Initially, we have 0.15 M of HNO2 and 0.20 M of NaNO2 in the solution. Since NaNO2 dissociates completely in water to release NO2-, we can consider the concentration of NO2- to be 0.20 M as well.
Now, using the expression for the acidic dissociation constant (Ka), we have:
Ka = [H3O+][NO2-]/[HNO2]
Substituting the given values:
4.0 × 10^-4 = [H3O+][0.20]/[0.15]
Simplifying, we get:
[H3O+] = (4.0 × 10^-4)(0.15)/0.20
[H3O+] ≈ 3.0 × 10^-4 M
Now, to find the pH, we use the formula:
pH = -log[H3O+]
pH ≈ -log(3.0 × 10^-4)
pH ≈ 3.52
Therefore, the pH of the aqueous solution containing 0.15 M HNO2 and 0.20 M NaNO2 is approximately 3.52.