For the diprotic weak acid H2A Ka1=2.1x10^-5 and Ka2=5.5x10^-7. What is the pH of a 0.0800M solution of H2A? What are the equilibrium concentrations of H2A and A^2- in this solution?

...........H2A ==> H^+ + HA^-

I.........0.0800....0.....0
C.........-x........x......x
E........0.0800-x...x.....x

k1 = (H^+)(HA^-)/(H2A)
Substitute and solve for x = (H^+) and convert to pH. Note that the k2 is about 100 times smaller than k1 so ignoring k2 doesn't cause much error.
I obtained 0.0787M for (H2A--you can round to correct sig figures) and for A^2- it is done this way.
k2 = 5.5E-7 = (H+)(A^=)/(HA^-). Since (H^+) = (HA^-), then (A=) = k2.

To find the pH of a 0.0800 M solution of H2A, we need to consider the ionization of the weak acid. H2A is a diprotic weak acid, which means it can donate two protons in separate stages. The first ionization constant, Ka1, is used to determine the equilibrium concentrations of H2A and H+ ions, while the second ionization constant, Ka2, is used to determine the equilibrium concentrations of HA- and H+ ions.

First, consider the ionization of H2A using Ka1:
H2A ⇌ H+ + HA-

The equilibrium equation for the ionization of H2A can be written as:
[H+][HA-] / [H2A] = Ka1

Let x be the equilibrium concentration of H+ ions and HA-. Since the initial concentration of H2A is 0.0800 M, the equilibrium concentration of H2A is (0.0800 - x) M. The equilibrium concentration of HA- is also x M.

Plugging these values into the equation and substituting the value of Ka1 (2.1 × 10^-5):
(x)(x) / (0.0800 - x) = 2.1 × 10^-5

Simplifying the equation, we have:
x^2 = 2.1 × 10^-5 (0.0800 - x)
x^2 = 1.68 × 10^-4 - 2.1 × 10^-5x

Rearranging the equation, we get:
x^2 + 2.1 × 10^-5x - 1.68 × 10^-4 = 0

Now, you can solve this quadratic equation for x using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 1, b = 2.1 × 10^-5, and c = -1.68 × 10^-4. Plugging in these values, we get:
x = (-2.1 × 10^-5 ± √((2.1 × 10^-5)^2 - 4(1)(-1.68 × 10^-4))) / (2(1))

Simplifying and solving the equation, you will obtain two values for x. Make sure to choose the appropriate value based on the context of the question.

Once you have the value of x, you can calculate the pH using the formula:
pH = -log[H+]

Regarding the equilibrium concentrations of H2A and A2- (HA-), you can determine them using the values of x obtained from the quadratic equation. The equilibrium concentration of H2A is (0.0800 - x) M, and the equilibrium concentration of HA- is x M.

To find the pH of the solution of H2A, we need to consider the dissociation of the weak acid H2A into its ions H+ and A^2-. Since H2A is diprotic, it has two ionization steps.

Step 1: H2A ⇌ H+ + HA^-
The equilibrium expression for this step is: Ka1 = [H+][HA-] / [H2A]

Step 2: HA^- ⇌ H+ + A^2-
The equilibrium expression for this step is: Ka2 = [H+][A^2-] / [HA-]

Given:
Ka1 = 2.1x10^-5
Ka2 = 5.5x10^-7

We will start by assuming that x is the concentration of H+ ions produced in the first step and in the second step.

For step 1:
[H+] = x
[HA-] = x
[H2A] = 0.0800 - x

Using the equilibrium expression for step 1, we can write:
Ka1 = [H+][HA-] / [H2A]
2.1x10^-5 = x*x / (0.0800 - x)

Solving this equation for x gives us the concentration of H+ ions after the first ionization step.

For step 2:
[H+] = x
[A^2-] = x
[HA-] = 0.0800 - x

Using the equilibrium expression for step 2, we can write:
Ka2 = [H+][A^2-] / [HA-]
5.5x10^-7 = x*x / (0.0800 - x)

Solving this equation for x gives us the concentration of H+ ions after the second ionization step.

To calculate the pH of the solution, we need to determine the concentration of H+ ions. Since H2A is a diprotic acid, there will be two equilibrium expressions and two equations to solve.