Calculate the concentrations of all species in a 0.320-M solution of H2X if K1 = 6.31E-06 and K2 = 3.71E-09
[H2X] =
[H3O1+] =
[HX1-] =
[X2-] =
Work:
H2X + H2O --> HX- + H3O+
.320 M
This is literally the only thing I know. How can there be two different Ks??
You started very well. Just go from there. There can be two Ka values because there are two H ions to dissociate.
.......H2X + H2O ==> H3O^+ + HX^-
I....0.320............0.......0
C.....-x..............x.......x
E...0.320-x...........x.......x
Now Ka1 = (H3O^+)(HX^-)/(H2X)
Substitute the E line into the Ka1 expression and solve for x = (H3O^+) = (HX^-). You get (H2X) from 0.320-x = ?
Then for ka2, it is
.....HX^- + H2O ==> H3O^+ + X^2-
You don't need to fill this in.
Ka2 = (H3O^+)(X^2-)/(HX^-)
You just solved from Ka1 that (H3O^+) = (HX^-). One of those is in the numerator of Ka2 and the other one in the denominator of Ka2 so they cancel which leaves Ka2 = (X^2-).
Case closed.
To solve this problem, we will use the concept of equilibrium constant expressions and the given equilibrium reaction. The presence of two different equilibrium constants, K1 and K2, suggests that the reaction undergoes two steps.
Let's break down the given reaction into its two steps:
Step 1: H2X + H2O ⇌ HX- + H3O+ (Equilibrium constant K1 = [HX-][H3O+]/[H2X][H2O])
Step 2: HX- + H2O ⇌ X2- + H3O+ (Equilibrium constant K2 = [X2-][H3O+]/[HX-][H2O])
Now, let's solve for the concentrations of all species in the solution:
Step 1:
Let's assume the concentration of H2O remains constant. Since the initial concentration of H2X is given as 0.320 M, we can represent the concentration of H2X as [H2X] = 0.320 M.
Now, using the equilibrium constant expression for step 1, substitute the given values and the concentration of H2X:
K1 = [HX-][H3O+]/[H2X][H2O]
6.31E-06 = [HX-][H3O+]/(0.320 M × [H2O])
The concentration of H2O is usually considered constant and its value is approximately 55.5 M. Therefore, we can substitute [H2O] = 55.5 M:
6.31E-06 = [HX-][H3O+]/(0.320 M × 55.5 M)
Step 2:
Now, let's solve for the concentrations of the species in step 2. Since this is a sequential reaction, the products of step 1 become the reactants of step 2.
Using the equilibrium constant expression for step 2, substitute the given values and the concentrations of species involved:
K2 = [X2-][H3O+]/[HX-][H2O]
3.71E-09 = [X2-][H3O+]/([HX-] × [H2O])
Next, we need to determine the concentration of [HX-]. Since it is a reactant in step 2, its concentration can be determined using the equilibrium constant expression obtained in step 1:
K1 = [HX-][H3O+]/(0.320 M × 55.5 M)
Solving for [HX-], we get:
[HX-] = K1 × (0.320 M × 55.5 M)/[H3O+]
Finally, substitute the obtained value of [HX-] in the equilibrium constant expression for step 2:
3.71E-09 = [X2-][H3O+]/(K1 × (0.320 M × 55.5 M)/[H3O+] × [H2O])
Now, you have two equations:
Equation 1: 6.31E-06 = [HX-][H3O+]/(0.320 M × 55.5 M)
Equation 2: 3.71E-09 = [X2-][H3O+]/(K1 × (0.320 M × 55.5 M)/[H3O+] × [H2O])
You can solve these two equations simultaneously to find the concentrations [H3O+], [HX-], [X2-], and [H2O].
Please note that the actual numerical values of the concentrations can only be determined by solving the system of equations involving these equilibrium constants and the given initial concentration of H2X.