1. 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.
To find the pH of a solution of nitrous acid (HNO2), we can use the equation for the dissociation of the acid and the equilibrium expression for the acid dissociation constant (Ka).
The dissociation of nitrous acid can be written as follows:
HNO2 ⇌ H+ + NO2-
The equilibrium expression for the acid dissociation constant (Ka) is:
Ka = [H+][NO2-] / [HNO2]
Given the Ka value of 4.5 × 10-4 and the molarity of the nitrous acid solution (0.085 M), we can assume that the concentration of H+ from the dissociation of nitrous acid is much smaller than the initial concentration of nitrous acid. Therefore, we can approximate the concentration of [H+] as x (the concentration of H+ that forms from the dissociation of nitrous acid) and the concentration of [NO2-] as x.
Substituting these approximations into the equilibrium expression for Ka:
4.5 x 10^-4 = (x)(x)/(0.085 - x)
We can neglect the x term in the denominator since x is small compared to the initial concentration of nitrous acid.
4.5 x 10^-4 = x² / 0.085
Rearranging the equation:
x² = (4.5 x 10^-4) * 0.085
x² = 3.825 x 10^-5
Taking the square root of both sides:
x = √(3.825 x 10^-5)
x = 0.0019563
Since we assumed that the concentration of H+ and NO2- is the same, the concentration of [H+] is approximately 0.0019563 M.
To find the pH, we use the equation:
pH = -log[H+]
pH = -log(0.0019563)
pH ≈ 2.708
Therefore, the pH of a 0.085 M solution of nitrous acid (HNO2) with a Ka of 4.5 × 10-4 is approximately 2.708. Answer choice 2.21 is the closest option to this value.
To find the pH of a solution of nitrous acid (HNO2), we can use the expression for the acid dissociation constant (Ka) and the equation for calculating the pH.
The balanced equation for the dissociation of nitrous acid is:
HNO2 ⇌ H+ + NO2-
Given that the concentration of nitrous acid (HNO2) is 0.085 M and the Ka value is 4.5 × 10^-4, we can set up an ICE table to represent the equilibrium concentrations:
Initial concentration: 0.085 M 0 M 0 M
Change in concentration: -x +x +x
Equilibrium concentration: 0.085 - x x x
The Ka expression for the equilibrium is:
Ka = [H+][NO2-] / [HNO2]
Substituting the equilibrium concentrations into the Ka expression, we get:
4.5 × 10^-4 = x^2 / (0.085 - x)
Since the value of x is expected to be much smaller than the initial concentration (0.085), we can approximate 0.085 - x as 0.085.
4.5 × 10^-4 = x^2 / 0.085
Simplifying the equation, we have:
4.5 × 10^-4 × 0.085 = x^2
3.825 × 10^-5 = x^2
Taking the square root of both sides, we find:
x ≈ 0.0062
Now that we know the concentration of H+ ions (x), we can use the equation for pH:
pH = -log[H+]
Substituting the value of x, we get:
pH = -log(0.0062)
Calculating this value, we find:
pH ≈ 2.21
Therefore, the pH of the 0.085 M solution of nitrous acid (HNO2) is approximately 2.21.