For a solution that is 0.280M HC3H5O2 (propionic acid, Ka= 1.3*10^-5) and 0.0894M HI , calculate the following:
[H3O+]
[OH-]
[C3H5O2-]
[I-]
You should write out the equations to understand what is going on. Let's call propionic acid HP.
.............HP ==> H+ + P-
initial.....0.280M...0....0
change......-x.......x.....x
equil.....0.280-x....x......x
...........HI ==> H+ + I-
initial..0.0894....0....0
equil.......0..0.0894..0.0894
Ka = (H+)(P-)/(HP)
Substitute from the ICE charts and solve for (P-). (H+) = 0.0894+x and (HP) = 0.280-x
To calculate the concentration of [H3O+], [OH-], [C3H5O2-], and [I-], we need to consider the dissociation reactions of the given compounds. Let's solve each step-by-step.
1. Dissociation of HC3H5O2 (propionic acid):
HC3H5O2 ⇌ H+ + C3H5O2-
The initial concentration of HC3H5O2 is 0.280 M. Assuming that the dissociation is complete, the concentration of H+ is equal to the concentration of HC3H5O2, which is 0.280 M.
[H3O+] = 0.280 M
2. Dissociation of HI:
HI ⇌ H+ + I-
The initial concentration of HI is 0.0894 M. Assuming that the dissociation is complete, the concentration of H+ is equal to the concentration of HI, which is 0.0894 M.
[H3O+] = 0.0894 M
3. Calculating [OH-]:
[H3O+] and [OH-] are related by the equation: [H3O+][OH-] = 1.0 * 10^-14 M^2 (at 25°C)
To calculate [OH-], we can rearrange the equation:
[OH-] = (1.0 * 10^-14 M^2) / [H3O+]
Using the value of [H3O+] from the previous calculations:
[OH-] = (1.0 * 10^-14 M^2) / 0.0894 M
[OH-] ≈ 1.12 * 10^-13 M
4. Calculating [C3H5O2-]:
The concentration of C3H5O2- is the same as the concentration of HC3H5O2 that dissociates. We assumed the dissociation is complete, so the concentration of C3H5O2- is equal to the initial concentration of HC3H5O2.
[C3H5O2-] = 0.280 M
5. Calculating [I-]:
The concentration of I- is the same as the concentration of HI that dissociates. We assumed the dissociation is complete, so the concentration of I- is equal to the initial concentration of HI.
[I-] = 0.0894 M
Summary of concentrations:
[H3O+] ≈ 0.0894 M
[OH-] ≈ 1.12 * 10^-13 M
[C3H5O2-] = 0.280 M
[I-] = 0.0894 M
To calculate the values of [H3O+], [OH-], [C3H5O2-], and [I-], we need to consider the reactions that occur between the reactants and their respective ions.
1. [H3O+]: Since HI is a strong acid and fully ionizes in water, we can assume that all the HI will dissociate to form H3O+ ions. Therefore, the concentration of [H3O+] will be equal to the initial concentration of HI, which is 0.0894 M.
2. [OH-]: To find the concentration of hydroxide ions, we need to consider the autoionization of water, which is a reversible reaction given by:
H2O ⇌ H+ + OH-
The concentration of [H+] and [OH-] in pure water is 1.0 x 10^-7 M each at 25°C. However, because propionic acid is a weak acid, there will be a small contribution of H+ ions from its partial dissociation. We can use the expression of the ionization constant (Ka) to calculate the [H+] concentration.
The dissociation of propionic acid (C3H5O2H) is given by:
C3H5O2H ⇌ C3H5O2- + H+
Using the expression for Ka = [C3H5O2-][H+]/[C3H5O2H], and noting that the initial concentration of propionic acid is 0.280 M, we can rearrange the equation to solve for [H+].
[H+] = (Ka * [C3H5O2H]) / [C3H5O2-]
[H+] = (1.3 x 10^-5 * 0.280) / 0.280
[H+] = 1.3 x 10^-5 M
Since water autoionization also contributes [H+] ions, the final concentration of [H+] will be the sum of the contributions from the acid and water autoionization:
[H3O+] = [H+] + [H2O]
[H3O+] = 1.3 x 10^-5 + 1.0 x 10^-7
[H3O+] = 1.31 x 10^-5 M
Using the equation for the ion product of water (Kw = [H+][OH-]), we can calculate [OH-]:
Kw = [H+][OH-]
1.0 x 10^-14 = (1.31 x 10^-5)([OH-])
[OH-] ≈ 7.63 x 10^-10 M
3. [C3H5O2-]: The concentration of the propionate ion can be obtained by assuming that all the propionic acid (C3H5O2H) ionizes. Therefore, the concentration of [C3H5O2-] will be equal to the initial concentration of propionic acid, which is 0.280 M.
[C3H5O2-] = 0.280 M
4. [I-]: Since HI is a strong acid, it will completely dissociate to form I- ions. Therefore, the concentration of [I-] will be equal to the initial concentration of HI, which is 0.0894 M.
[I-] = 0.0894 M
To summarize:
[H3O+] ≈ 1.31 x 10^-5 M
[OH-] ≈ 7.63 x 10^-10 M
[C3H5O2-] = 0.280 M
[I-] = 0.0894 M