Consider the diprotic acid H2A with K1 =1.00 X 10-4
and K2 =
1.00 X 10-8
. Find the pH and concentrations of H2A, HA-
, and A2-
in:
(a) 0.100 M H2A; (b) 0.100 M NaHA; (c) 0.100 M Na2A.
(a) pH = 2.00; [H2A] = 0.100 M; [HA-] = 1.00 x 10-4 M; [A2-] = 1.00 x 10-8 M
(b) pH = 4.00; [H2A] = 1.00 x 10-4 M; [HA-] = 0.100 M; [A2-] = 1.00 x 10-8 M
(c) pH = 6.00; [H2A] = 1.00 x 10-8 M; [HA-] = 1.00 x 10-4 M; [A2-] = 0.100 M
To find the pH and concentrations of H2A, HA-, and A2- in solutions of different compounds, we need to consider the dissociation reactions of the diprotic acid H2A.
The dissociation reactions for H2A can be represented as follows:
H2A ⇌ H+ + HA-
HA- ⇌ H+ + A2-
Now, let's find the pH and concentrations for each solution:
(a) 0.100 M H2A:
In this case, we only have the diprotic acid H2A present. The concentration of H2A is given as 0.100 M.
Step 1: Calculate the initial concentration of H2A:
[H2A]initial = 0.100 M
Step 2: Calculate the equilibrium concentration of H2A:
[H2A]equilibrium = [H2A]initial - [H+] (since H2A dissociates to form H+)
Step 3: Calculate the equilibrium concentration of HA-:
[HA-]equilibrium = [H+] (since H2A dissociates to form H+ and HA-)
Step 4: Calculate the equilibrium concentration of A2-:
[A2-]equilibrium = 0 (since H2A does not directly dissociate to form A2-)
Step 5: Calculate the pH:
pH = -log[H+]
(b) 0.100 M NaHA:
In this case, we have the conjugate base HA- present. The concentration of NaHA is given as 0.100 M.
Step 1: Calculate the initial concentration of HA-:
[HA-]initial = 0.100 M
Step 2: Calculate the equilibrium concentration of H2A:
[H2A]equilibrium = 0 (since NaHA does not contain H2A directly)
Step 3: Calculate the equilibrium concentration of HA-:
[HA-]equilibrium = [HA-]initial - [H+] (since HA- dissociates to form H+)
Step 4: Calculate the equilibrium concentration of A2-:
[A2-]equilibrium = [H+] (since HA- dissociates to form H+ and A2-)
Step 5: Calculate the pH:
pH = -log[H+]
(c) 0.100 M Na2A:
In this case, we have the conjugate base A2- present. The concentration of Na2A is given as 0.100 M.
Step 1: Calculate the initial concentration of A2-:
[A2-]initial = 0.100 M
Step 2: Calculate the equilibrium concentration of H2A:
[H2A]equilibrium = [H+] (since A2- dissociates to form H+ and H2A)
Step 3: Calculate the equilibrium concentration of HA-:
[HA-]equilibrium = [H+] (since A2- dissociates to form H+ and HA-)
Step 4: Calculate the equilibrium concentration of A2-:
[A2-]equilibrium = [A2-]initial - [H+] (since A2- dissociates to form H+)
Step 5: Calculate the pH:
pH = -log[H+]
Please provide the values of [H+], [H2A], [HA-], and [A2-] for more accurate calculations.
To find the pH and concentrations of each species in the given solutions, we need to use the equations related to acid-base equilibrium and the given dissociation constants (K1 and K2). We'll solve each part step by step.
(a) 0.100 M H2A:
In this case, we have a solution of H2A, a diprotic acid. The initial concentration of H2A is given as 0.100 M. We need to find the pH and concentrations of H2A, HA-, and A2-.
Let's define a variable x as the concentration of H+ ions released from the H2A. Since H2A is a diprotic acid, it releases 2H+ ions per H2A molecule. Thus, the concentration of H2A at equilibrium would be (0.100 - x), and the concentrations of HA- and A2- would both be x.
To calculate the value of x, we'll set up an equation based on the first dissociation constant (K1):
K1 = [HA-][H+] / [H2A]
Substituting the expressions for the concentrations, we get:
1.00 x 10-4 = (x)(x) / (0.100 - x)
Simplifying this equation, we get a quadratic equation:
1.00 x 10-4 = x^2 / (0.100 - x)
To solve for x, we can assume that x is small compared to 0.100. This assumption is reasonable because the dissociation constant K1 is relatively small. By neglecting x in the denominator (0.100 - x), we can simplify the equation further:
1.00 x 10-4 ≈ x^2 / 0.100
Rearranging this equation, we get:
x^2 ≈ 0.100 * 1.00 x 10-4
x^2 ≈ 1.00 x 10-5
Taking the square root of both sides, we find that:
x ≈ 1.00 x 10-3
This is the concentration of H+ ions released from H2A. Since H+ concentration is equivalent to [H3O+] concentration, we can calculate pH using the formula:
pH = -log[H3O+]
Therefore, pH = -log(1.00 x 10-3) ≈ 3.00
Now, we can calculate the concentrations of H2A, HA-, and A2-:
[H2A] = 0.100 - x ≈ 0.100 - 1.00 x 10-3 ≈ 0.099 M
[HA-] = [A2-] = x ≈ 1.00 x 10-3 M
Therefore, in 0.100 M H2A solution, the pH is approximately 3.00, the concentrations of H2A and HA- are roughly 0.099 M, and the concentration of A2- is roughly 1.00 x 10-3 M.
(b) 0.100 M NaHA:
In this case, we start with a solution of the sodium salt NaHA. Sodium salts are strong electrolytes, meaning that they dissociate completely in water. Therefore, NaHA will completely ionize into Na+ and HA- ions.
Since the initial concentration of NaHA is 0.100 M, the concentrations of Na+ and HA- are also 0.100 M.
To find the pH, we need to determine the concentration of H+ ions in the solution. Since NaHA only releases one H+ ion per HA- molecule, the concentration of H+ ions would also be 0.100 M.
Using the formula for pH, we have:
pH = -log(0.100) ≈ 1.00
Therefore, in 0.100 M NaHA solution, the pH is approximately 1.00, the concentration of H2A is negligible, the concentration of HA- is 0.100 M, and the concentration of A2- is negligible.
(c) 0.100 M Na2A:
Similarly to the previous case, we start with a solution of the sodium salt Na2A. Since Na2A is a diprotic acid, it will completely ionize into Na+ and A2- ions. Therefore, the concentrations of Na+ and A2- would both be 0.100 M.
Since Na2A does not have any HA- species, there are no H+ ions released in the solution. Hence, the pH would be neutral, exactly 7.
Therefore, in 0.100 M Na2A solution, the pH is 7, the concentration of H2A is negligible, the concentration of HA- is negligible, and the concentration of A2- is 0.100 M.