What is the pH of a 0.085 M solution of nitrous acid (HNO2) that has a Ka of 4.5 × 10^-4?
2.21
5.33
3.35
4.42
I am not the best at working through the problems, but I came to A, 2.21, given the concentration and Ka.
HNO2 <> H+ + NO2-
ka=x^2/(.085-x)
.085ka-ka*x=x^2
x^2+4.5*x*e-4 -.085*4.5e-4=0 >>> http://www.math.com/students/calculators/source/quadratic.htm
x=0.0059637498737628745
p(x)=2.22448058
Great, thank you for confirming my answer!!
Is this K12 AP Chemistry? I need some help myself on the matter.
To find the pH of the solution, we can use the information given about the concentration (0.085 M) and the Ka value (4.5 × 10^-4) of nitrous acid (HNO2).
Step 1: Set up the equation for the dissociation of nitrous acid (HNO2):
HNO2 ⇌ H+ + NO2-
Step 2: Write the expression for the Ka value:
Ka = [H+][NO2-]/[HNO2]
Step 3: Substitute the given values into the Ka expression:
4.5 × 10^-4 = [H+][NO2-]/0.085
Step 4: Since the [NO2-] concentration is negligible compared to [HNO2], we can assume that the concentration of H+ is equal to [HNO2].
4.5 × 10^-4 = [H+]^2/0.085
Step 5: Solve for [H+]:
[H+]^2 = 4.5 × 10^-4 × 0.085
[H+]^2 = 3.825 × 10^-5
[H+] = √(3.825 × 10^-5)
[H+] ≈ 0.00619 M
Step 6: Calculate the pH using the formula: pH = -log[H+]
pH ≈ -log(0.00619)
pH ≈ 2.21
Therefore, the correct answer is A, 2.21.