Determine the concentration of CO32- ions in a 0.18 M H2CO3 solution. Carbonic acid is a diprotic acid whose Ka1 = 4.3 × 10-7 and Ka2 = 5.6 × 10-11.
(CO3^2-) = k2. The reasoning is this but it's a little long. I wish we had a board and I could talk.
...........H2CO3 ==> H^+ + HCO3^-
I..........0.18......0.......0
C...........-x.......x.......x
E.........0.18-x.....x.......x
k1 = 4.3E-7 = (x)(x)/(0.18-x) and solve for x = (H^+) = (HCO3^-)if you want an answer but that really isn't necessary. That answer is about 2.8E-4M. Then k2 takes over.
...........HCO3^- ==> H^+ + CO3^2-
I..........2.8E-4....2.8E-4...0
C..........-x.........x.......x
E.........2.8E-4-x..2.8E-4+x..x
so k2 = (H^+)(HCO3^-)/(HCO3^-). If you substitute the numbers you have
k2 = (2.8E-4+x)(x)/(2.8E-4-x)
if x is small the k2 = x = (CO3^2-)
As above, however, you really don't need to solve for H^+ and HCO3^-. Your reasoning tells you (H^+) = (HCO3^-) and since k2 = (H^+)(CO3^2-)/(HCO3^-),they cancel each other in the k2 equation and (CO3^2-) = k2
Well, I have to say, this question is quite a carbonated one! But don't worry, I'm here to carbonate... I mean clarify things for you!
Since carbonic acid is a diprotic acid, it can donate two protons (H+) in solution. The first protonation reaction gives us bicarbonate ion (HCO3-), and the second protonation reaction gives us carbonate ion (CO32-).
What we need to do is find the concentration of CO32- ions in the solution. To do this, we first need to find the concentration of bicarbonate ions (HCO3-) using the given information.
Let's start by assuming x is the concentration of HCO3-. When one proton is donated from carbonic acid, it forms the bicarbonate ion (HCO3-). This means that the concentration of bicarbonate ions (x) will also be the concentration of H+ ions in solution.
Using the first dissociation constant (Ka1) of 4.3 × 10-7 and the equation:
Ka1 = [HCO3-][H+]/[H2CO3]
Since the concentration of H2CO3 is given as 0.18 M and we know [HCO3-] = [H+], we can substitute the values into the equation:
4.3 × 10-7 = (x)(x)/(0.18)
Now, let's solve for x:
x^2 = 4.3 × 10-7 × 0.18
x^2 = 7.74 × 10-8
x ≈ 2.78 × 10-4 M
So, the concentration of bicarbonate ions (HCO3-) in the solution is approximately 2.78 × 10-4 M. And since the concentration of carbonate ions (CO32-) is equal to the concentration of bicarbonate ions (HCO3-), we can say that the concentration of CO32- ions is also approximately 2.78 × 10-4 M.
I hope that cleared things up for you! Carbonation, I mean, chemistry can be bubbly at times, but it's all part of the fun!
To determine the concentration of CO32- ions in the H2CO3 solution, we need to understand the dissociation of carbonic acid (H2CO3) and the subsequent formation of CO32- ions.
H2CO3 dissociates in two steps:
Step 1: H2CO3 ⇌ H+ + HCO3-
- This reaction has a dissociation constant (Ka1) of 4.3 × 10-7.
Step 2: HCO3- ⇌ H+ + CO32-
- This reaction has a dissociation constant (Ka2) of 5.6 × 10-11.
Let's denote the initial concentration of H2CO3 as [H2CO3]_initial and the final concentration as [H2CO3]_final. We also assume that the concentrations of H+ and CO32- are x, which is also the concentration of HCO3- formed from the first step.
Using the dissociation constants, we can set up the following equations:
Ka1 = [H+][HCO3-] / [H2CO3]_initial
Ka2 = [H+][CO32-] / [HCO3-]
We can see that [HCO3-] = x and [H2CO3]_initial = [H2CO3]_final, since H2CO3 is a weak acid and does not dissociate significantly.
Since the concentration of H+ is the same in both equations, we can equate the two equations:
Ka1 = x^2 / [H2CO3]_final
Ka2 = [CO32-] * x / x
Solving for x, we find:
x = sqrt(Ka1 * [H2CO3]_final)
Substituting the given values, we have:
x = sqrt((4.3 × 10-7) * (0.18 M))
Evaluating this, we get:
x ≈ 2.99 × 10-4 M
Therefore, the concentration of CO32- ions in a 0.18 M H2CO3 solution is approximately 2.99 × 10-4 M.
To determine the concentration of CO32- ions in a 0.18 M H2CO3 solution, we need to consider the dissociation of carbonic acid (H2CO3) and the corresponding equilibrium constants (Ka1 and Ka2).
The dissociation of carbonic acid can be represented by the following reactions:
H2CO3 ⇌ H+ + HCO3- (Equation 1)
HCO3- ⇌ H+ + CO32- (Equation 2)
The equilibrium constant Ka1 represents the dissociation of the first hydrogen ion from carbonic acid (Equation 1).
The equilibrium constant Ka2 represents the dissociation of the second hydrogen ion from bicarbonate ion (Equation 2).
We can start by calculating the concentration of H+ ions produced from the dissociation of carbonic acid using the first equilibrium constant (Ka1).
Ka1 = [H+][HCO3-] / [H2CO3]
Since the concentration of H+ ions is equal to the concentration of HCO3- ions (as per Equation 1), we can substitute [HCO3-] with the x symbol, where x represents the concentration of HCO3-.
Therefore:
Ka1 = x^2 / (0.18 - x)
Now, let's solve for x (the concentration of H+ and HCO3- ions) by substituting the given value of Ka1 (4.3 × 10^-7):
4.3 × 10^-7 = x^2 / (0.18 - x)
We can neglect the value of x in the denominator since it is significantly smaller compared to 0.18. By doing this approximation, the equation becomes:
4.3 × 10^-7 = x^2 / 0.18
To solve for x, we rearrange the equation:
x^2 = (4.3 × 10^-7) * 0.18
x = √[(4.3 × 10^-7) * 0.18]
Solving these calculations, we find that x ≈ 8.35 × 10^-4 M
Now, this concentration value represents both the concentration of H+ and HCO3- ions. However, we need to find the concentration of CO32- ions specifically.
Since the concentration of HCO3- ions is equal to the concentration of CO32- ions (as per Equation 2), we can directly conclude that the concentration of CO32- ions is also 8.35 × 10^-4 M.
Therefore, the concentration of CO32- ions in the 0.18 M H2CO3 solution is approximately 8.35 × 10^-4 M.