in a 0.40 M solution of diprotic acid H2A(Ka=7.4x10^-5, Ka2=5.0x10^-10 at 25 celsius), what is the equilibrium concentration of A^2-? A)0.40m b)0.80M C)5.4x10^-3m d)1.4x10^-5M E)5.0x10^-10M
H2A+ H2O----H3O+ HA^-2
HA^-2+ H2O-----> H3O + A-
Let x=HA^-2
Let y=A-
So, for the following reaction
H2C2O4 + H2O----H3O+ HC2O4
H2C2O4...........H3O....HC2O4
I0.40M..............0 ........ 0
C-x.....................x............ x
E0.40-x.............x.............x
Solve for x,
ka1=[x][x]/[0.40-x], which turns into,
ka1=[x][x]/[0.40]
sqrt*(ka1*0.40)=x=5.4 x 10^-3
HC2O4 + H2O-----> H3O + C2O4
HC2O4...........H3O....C2O4
I.......Y..............0 ............ 0
C....-z...............z.............z
E....y-z..............z.............z
But since it is a diprotic acid and x=H3O+ y= H3O+, the charts become
H2C2O4 + H2O----H3O+ HC2O4
H2C2O4...........H3O....HC2O4
I0.4M..............0 ........ ..0
C-x..............,,..x............ .x
E0.4-x.........x+Y..........x-y
HC2O4 + H2O-----> H3O + C2O4
HC2O4..................H3O...............C2O4
I...5.4 x 10^-3.........x.....................0
C..-y......................x+y.....................y
E...5.4 x 10^-3-y...x+y.....................y
So, ka1=[5.4 x 10^-3+y][5.4 x 10^-3-y]/[0.370]
and
ka2=[5.4 x 10^-3+y][y]/[5.4 x 10^-3]
We solved for x, so solve for y.
I believe this is how you tackle this problem.
See the above.
Change H2C2O4 to H2A, HC2O4 to HA^2, and C2O4 to A-. Everything else is okay.
To find the equilibrium concentration of A^2- in a 0.40 M solution of diprotic acid H2A, we need to consider the dissociation of the acid and the equilibrium expressions for the acid dissociation constants (Ka values).
Let's start by writing the dissociation reactions:
H2A ⇌ H+ + HA^- (Equation 1)
HA^- ⇌ H+ + A^2- (Equation 2)
Since H2A is a diprotic acid, it can lose two protons (H+ ions) in separate dissociation steps.
Now, let's define some variables:
[H2A] = Initial concentration of H2A = 0.40 M
[H+] = Concentration of H+ ions
[HA^-] = Concentration of HA^- ions
[A^2-] = Concentration of A^2- ions
Now we can write the equilibrium expressions for the dissociation reactions:
Ka1 = [H+][HA^-] / [H2A]
Ka2 = [H+][A^2-] / [HA^-]
We know that Ka1 = 7.4x10^-5 and Ka2 = 5.0x10^-10.
Considering Equation 1, we can assume that [H+] ≈ [HA^-] since H2A is a weak acid. Therefore, we can rewrite Ka1 expression as:
Ka1 = [H+]^2 / [H2A]
Using the given Molarity for H2A (0.40 M), we can calculate [H+]:
7.4x10^-5 = [H+]^2 / 0.40
[H+]^2 = 7.4x10^-5 * 0.40
[H+]^2 = 2.96x10^-5
[H+] = √(2.96x10^-5)
[H+] ≈ 5.44x10^-3 M
Since [H+] ≈ [HA^-], we can approximate [HA^-] as 5.44x10^-3 M.
Now, using the equilibrium expression for Equation 2:
Ka2 = [H+][A^2-] / [HA^-]
We plug in the values:
5.0x10^-10 = (5.44x10^-3)([A^2-]) / (5.44x10^-3)
Solving for [A^2-], we get:
5.0x10^-10 * 5.44x10^-3 = [A^2-]
[A^2-] ≈ 2.72x10^-12 M
Therefore, the equilibrium concentration of A^2- in the 0.40 M solution of diprotic acid H2A is approximately 2.72x10^-12 M.
The correct option that represents this concentration is answer D) 1.4x10^-5 M.