Suppose that 2.00 mol of HNO2 is dissolved in 1.00L of water at 25 degrees C. Because HNO2 is a weak acid, it is in equilibrium with the ions, H+ and NO2-. What are the concentrations of these ions at equilibrium?
To find the concentrations of H+ and NO2- ions at equilibrium, we need to consider the dissociation of HNO2.
The balanced equation for the dissociation of HNO2 is:
HNO2 ⇌ H+ + NO2-
The equilibrium expression for this reaction is:
K = [H+][NO2-]/[HNO2]
In this case, we're given that 2.00 mol of HNO2 is dissolved in 1.00L of water.
The concentration of HNO2 can be calculated by dividing the amount of HNO2 by the volume of the solution:
[HNO2] = moles of HNO2 / volume of solution
[HNO2] = 2.00 mol / 1.00 L = 2.00 M
Since H+ and NO2- are formed in a 1:1 ratio, the concentration of H+ and NO2- ions at equilibrium will be the same.
Let's assume the concentration of H+ and NO2- ions at equilibrium is x M.
Substituting the given values into the equilibrium expression, we get:
K = [H+][NO2-]/[HNO2]
K = (x)(x)/(2.00)
Given that K is the equilibrium constant, we need its numerical value to solve for x.
Please provide the value of the equilibrium constant K for this reaction.
To determine the concentrations of H+ and NO2- ions at equilibrium, we need to use the equilibrium constant expression (Ka) for the dissociation of HNO2.
The balanced chemical equation for the dissociation of HNO2 is:
HNO2 ⇌ H+ + NO2-
The equilibrium constant expression (Ka) for this reaction can be written as:
Ka = [H+][NO2-] / [HNO2]
Given that 2.00 mol of HNO2 is dissolved in 1.00L of water, we can use the concentration (molarity) to calculate the initial concentration of HNO2:
[HNO2]initial = 2.00 mol / 1.00 L = 2.00 M
At equilibrium, let's assume the concentration of H+ and NO2- be x (in mol/L). So, we can write the concentrations as:
[H+] = x
[NO2-] = x
Now, substitute these values into the equilibrium constant expression:
Ka = [H+][NO2-] / [HNO2]
Ka = x * x / 2.00
Now, we need the value of Ka for HNO2. If it is not given, we need to look it up in a reliable source such as a chemistry textbook or online database. For the sake of example, let's assume the Ka value is 4.5 x 10^-4 (hypothetical value).
Plugging in the values, we get:
4.5 x 10^-4 = x * x / 2.00
Now, we can solve this equation for x.
Rearranging the equation, we get:
x * x = 4.5 x 10^-4 * 2.00
x^2 = 9.0 x 10^-4
Taking the square root of both sides, we get:
x ≈ 0.03
So, the concentration of both H+ and NO2- ions at equilibrium is approximately 0.03 mol/L.