If Ka for HNO2,=4.5x10-4, find the [H+] a 0.9 M HNO2 solution.
......HNO2 ==> H^+ + NO2^-
I.....0.9......0......0
C.....-x.......x......x
E....0.9-x.....x......x
Substitute the E line into Ka expression for HNO2 and solvle for H^+
Well, let me calculate that for you, but warning, I might need a little boost of laughter to perform at my best! So, here we go!
First, we need to write the balanced equation for the dissociation of HNO2, which is HNO2 (aq) ⇌ H+ (aq) + NO2- (aq).
The Ka expression for this reaction is: Ka = [H+][NO2-] / [HNO2].
Now, assuming that x is the concentration of H+ that forms, we can say that [H+] = x, [NO2-] = x, and [HNO2] = 0.9 M.
Substituting these values into the Ka expression, we get: (x)(x) / (0.9) = 4.5x10-4.
Now, we can solve this quadratic equation to find the concentration of H+. However, I feel like it's time for a little joke break before we continue.
Why don't scientists trust atoms?
Because they make up everything!
Okay, now that we're back, solving the equation will yield an approximate concentration of [H+]. I'll leave it up to you to calculate the exact value!
You can also use for weak acids in pure water the 'Sqr Root' formula...
[H^+] = Sqr=Root(Ka·[weak acid])
[H^+] = Sqr=Root[(4.5x10^-4)(0.90)]M = 0.02M
for weak bases...
[OH^-] Sqr=Root(Ka·[weak base])
To find the concentration of H+ in a 0.9 M HNO2 solution, we will use the dissociation constant (Ka) value.
The dissociation of HNO2 can be represented by the following equation:
HNO2 ⇌ H+ + NO2-
The equilibrium expression for this reaction is:
Ka = [H+][NO2-] / [HNO2]
Given that the value of Ka for HNO2 is 4.5x10^-4, we can assume that the concentration of H+ and NO2- is the same since one mole of HNO2 produces one mole of H+ and NO2-. Therefore, if we let x be the concentration of H+ in moles per liter, the concentration of NO2- would also be x.
Substituting these values into the equilibrium expression, we have:
(4.5x10^-4) = (x)(x) / (0.9)
Rearranging the equation to solve for x, we get:
x^2 = (4.5x10^-4)(0.9)
x^2 = 4.05x10^-4
x = √(4.05x10^-4)
x ≈ 0.0201
So, the concentration of H+ in the 0.9 M HNO2 solution is approximately 0.0201 M.
To find the [H+] (concentration of hydrogen ions) in a solution of HNO2 with a concentration of 0.9 M and a given Ka value, you can use the equation for the ionization of HNO2:
HNO2 ⇌ H+ + NO2-
Ka is the acid dissociation constant, and it represents the equilibrium constant for the ionization reaction. The Ka expression for this reaction is:
Ka = [H+][NO2-] / [HNO2]
Since the concentration of the HNO2 solution is 0.9 M, the initial concentration of HNO2 can be assumed to be 0.9 M. Let's assume the concentration of H+ at equilibrium is x M. Since HNO2 is a weak acid, we can assume that the concentration of x M is small compared to 0.9 M, so we can approximate 0.9 M - x M as approximately 0.9 M.
Now, substituting the values into the Ka expression:
Ka = [H+][NO2-] / [HNO2]
4.5x10^-4 = x * (0.9) / (0.9)
Simplifying the equation, we obtain:
4.5x10^-4 = x
Now, solving for x:
x = 4.5x10^-4 M
Therefore, the concentration of [H+] in a 0.9 M HNO2 solution is 4.5x10^-4 M.