Calculate the concentration of all species in a 0.130M solution of H2CO3
Enter your answers numerically separated by commas. Express your answer using two significant figures.
[H2CO3], [HCO?3], [CO2?3], [H3O+], [OH?] =
I presume you have [H2CO3] and [H2O^+].
You can use [H3O^+]{OH^-] = Kw = 1E-14 to solve for (OH^-).
(HCO3^-) = essentially the same as (H3O^+)
(CO3^2-) = Ka2
To calculate the concentration of all the species in a 0.130M solution of H2CO3, we need to consider the dissociation of H2CO3 in water. H2CO3 can dissociate into three different species: HCO3-, CO32-, and H3O+.
The balanced chemical equation for the dissociation of H2CO3 is as follows:
H2CO3 ⇌ HCO3- + H3O+
Since this is a weak acid, we can assume that the concentration of H2CO3 that dissociates is negligible compared to the initial concentration. Therefore, we can say that the concentration of H2CO3 is approximately equal to its initial concentration.
[H2CO3] = 0.130 M (since it does not dissociate significantly)
Now, let's calculate the concentrations of the other species using the equilibrium constant expression for the dissociation reaction:
[HCO3-][H3O+] / [H2CO3] = K
The equilibrium constant (K) for the dissociation of H2CO3 is approximately 1.77 x 10^-4 at 25°C.
Using this information, we can solve for [H3O+] to find the concentration of the remaining species.
Let's set up the calculation:
[HCO3-][H3O+] / [H2CO3] = K
[HCO3-][H3O+] / 0.130 = 1.77 x 10^-4
Now, rearrange the equation to solve for [H3O+]:
[H3O+] = (K * 0.130) / [HCO3-]
To find the concentration of [HCO3-], we need to make an assumption based on the pH of the solution. If we assume the solution is neutral (pH = 7), [H3O+] will be equal to [OH-]. Since [OH-] in pure water is 1 x 10^-7 M, we can make this assumption.
Now, let's calculate:
[H3O+] = (1.77 x 10^-4 * 0.130) / 1 x 10^-7
[H3O+] ≈ 0.229 M (rounded to two significant figures)
Since the solution is neutral, [OH-] will also be 0.229 M.
To calculate the concentrations of [HCO3-] and [CO32-], we can use the fact that [H2CO3] = [HCO3-] + [CO32-].
[HCO3-] + [CO32-] = 0.130 M
Since the initial concentration of H2CO3 is equal to the concentration of HCO3- and CO32-, we get:
[HCO3-] ≈ 0.130 - 0.229 ≈ 0.090 M
[CO32-] ≈ 0.090 M
Therefore, the concentrations of the species are:
[H2CO3] ≈ 0.130 M
[HCO3-] ≈ 0.090 M
[CO32-] ≈ 0.090 M
[H3O+] ≈ 0.229 M
[OH-] ≈ 0.229 M
Please note that these values are approximations rounded to two significant figures.
To find the concentrations of all the species in a 0.130M solution of H2CO3, we need to consider the dissociation equilibrium of H2CO3:
H2CO3 <--> HCO3- + H3O+
Given that the initial concentration of H2CO3 is 0.130M, we can assume that it fully dissociates, and the concentration of H2CO3 becomes 0M. So, the concentration of HCO3- and H3O+ will be equal to 0.130M.
Therefore:
[H2CO3] = 0M
[HCO3-] = 0.130M
[CO2-3] = 0M
[H3O+] = 0.130M
[OH-] = 0M
Thus, the concentrations of the species are:
[H2CO3], [HCO3], [CO2], [H3O+], [OH-] = 0M, 0.130M, 0M, 0.130M, 0M