A solution containing 0.0158 M maleic acid and 0.0226 M disodium maleate. The Ka values for maleic acid are 1.20 × 10-2 (Ka1) and 5.37 × 10-7 (Ka2). What is the pH of the solution ?

Well, with that combination, I'm sure the solution feels quite male-icious. But don't worry, I'm here to calculate the pH for you!

To solve this, we need to consider the dissociation of maleic acid and the hydrolysis of disodium maleate. Maleic acid dissociates into hydrogen ions (H+) and the maleate ion (C4H3O4-). Meanwhile, disodium maleate reacts with water to produce hydroxide ions (OH-) and the maleate ion.

Now, let's calculate the pH step by step. First, we need to find the concentrations of the individual ions.

For maleic acid (C4H4O4):
[H+] = [C4H3O4-] (because maleic acid is a weak acid)

For disodium maleate (C4H2Na2O4):
[OH-] = [C4H3O4-] (because disodium maleate hydrolyzes to produce OH- ions)

Now, let's calculate the concentrations:

[H+] = [C4H3O4-] = 0.0158 M (given)
[OH-] = 0.0226 M (given)

To calculate the pH, we need to use the equation: pH = -log[H+]. So let's plug in the value:

pH = -log(0.0158)
pH = 1.80

So, the pH of the solution is approximately 1.80. Remember, I'm just a clown bot, so don't clown around with this answer.

To find the pH of the solution, we need to determine the dominant form of the maleic acid and its corresponding Ka value.

The molecular formula of maleic acid is C4H4O4, and its structure can be represented as:

HOOC-CH=CH-COOH

Maleic acid can dissociate in water to release a proton (H+) due to its carboxylic acid groups. This results in the formation of two equilibrium reactions:

1) Maleic acid (H2MA) donating its first proton:
H2MA ⇌ H+ + MA-

2) The monoanion form (MA-) donating its second proton:
MA- ⇌ H+ + M2-

To determine which form of maleic acid will be the dominant form, we need to compare the initial concentrations of maleic acid and disodium maleate.

Given:
[H2MA] = 0.0158 M (concentration of maleic acid)
[MA-] = 0.0226 M (concentration of disodium maleate)

Since maleic acid donates protons and disodium maleate accepts protons, the reaction between them can be represented as:

H2MA + MA- ⇌ H+ + M2-

As the concentration of disodium maleate ([MA-]) is higher than that of maleic acid ([H2MA]), the reaction will shift towards the right (forward) to consume the excess MA- and produce more protonated forms.

Therefore, the dominant form will be MA-, and its corresponding Ka value (Ka2 = 5.37 × 10-7) will be used.

Now, let's set up an ice table to solve for the concentration of H+ and calculate the pH of the solution.

Step 1: Set up the initial concentrations:
[H2MA] = 0.0158 M
[MA-] = 0.0226 M
[H+] = 0 M (initial concentration of proton)

Step 2: Set up the change in concentration (x) due to the reaction:
[H2MA] decreases by x
[MA-] decreases by x
[H+] increases by x

Step 3: Calculate the new concentrations:
[H2MA] = 0.0158 - x
[MA-] = 0.0226 - x
[H+] = x

Step 4: Write the expression for Ka2:
Ka2 = [H+][M2-] / [MA-]

Since [H+] = x and [MA-] = 0.0226 - x, the expression becomes:
5.37 × 10-7 = x^2 / (0.0226 - x)

Step 5: Assume the approximation x << 0.0226, then we can neglect x compared to 0.0226:
5.37 × 10-7 = x^2 / 0.0226

Step 6: Solve for x:
x^2 = 0.0226 * 5.37 × 10-7
x^2 ≈ 1.21 × 10-8
x ≈ √(1.21 × 10-8)
x ≈ 1.10 × 10-4

Step 7: Calculate the pH:
pH = -log[H+]
pH = -log(1.10 × 10-4)
pH ≈ 3.96

Therefore, the pH of the solution containing 0.0158 M maleic acid and 0.0226 M disodium maleate is approximately 3.96.

To determine the pH of the solution, we need to consider the dissociation of the maleic acid and its conjugate base, the disodium maleate.

Maleic acid (HA) is a weak acid that undergoes dissociation in water according to the following equilibrium reaction:
HA ⇌ H+ + A-

The Ka1 value of 1.20 × 10-2 represents the equilibrium constant for the dissociation of the first H+ ion from maleic acid.

Disodium maleate (Na2A) is the conjugate base of maleic acid. It can accept H+ ions in water to form hydroxide ions (OH-) according to the following equilibrium reaction:
Na2A + H2O ⇌ 2Na+ + A- + OH-

To solve for the pH of the solution containing 0.0158 M maleic acid and 0.0226 M disodium maleate, we need to calculate the concentrations of H+ and OH- ions.

Step 1: Calculate the concentrations of H+ and A- from the dissociation of maleic acid:
- Since maleic acid and disodium maleate are in the same solution, we can assume that the initial concentration of maleic acid equals the concentration of H+ ions: [H+] = 0.0158 M.
- Since maleic acid and disodium maleate are in the same solution, we can assume that the initial concentration of disodium maleate equals the concentration of A- ions: [A-] = 0.0226 M.

Step 2: Calculate the concentrations of OH- ions:
- Since Na2A can accept H+ ions to form hydroxide ions, the concentration of OH- ions can be calculated using the concentration of A- ions: [OH-] = [A-].

Step 3: Calculate the concentration of water:
- In water, the product of [H+] and [OH-] is always equal to 1.0 x 10^-14 (at 25°C).
- Therefore, [H+] x [OH-] = 1.0 x 10^-14.
- Substitute the values: (0.0158) x (0.0226) = 1.0 x 10^-14.
- Rearrange the equation to solve for [OH-]: [OH-] = (1.0 x 10^-14) / (0.0158).

Step 4: Calculate the pOH:
- The pOH is calculated using the concentration of OH-: pOH = -log [OH-].

Step 5: Calculate the pH:
- The pH can be determined using the equation: pH = 14 - pOH.

By following these steps, you should be able to calculate the pH of the solution containing 0.0158 M maleic acid and 0.0226 M disodium maleate.