In a .25 M solution, a weak acid is 3.0% dissociated.

A. calculate the pH of the solution
B. calculate the Ka of the acid

Please explain steps.

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

initial....0.25M....0......0
change.......-x......x......x
equil......0.25-x....x.....x
If it is 3% dissociated, that means that x = 0.25*0.03 = ?
Substitute into the Ka expression and solve for Ka.
pH = -log(H^+)

Please help me solved this problem

Sure, let's break down the steps to solve the problem:

A. To calculate the pH of the solution:
1. Find the concentration of the acid after dissociation. Since the acid is 3.0% dissociated in the solution, this means that only 3.0% of the initial concentration remains in the solution.
Concentration of the acid after dissociation = 0.03 * 0.25 M = 0.0075 M

2. Use the concentration of the acid to calculate the pOH, which can be converted to pH.
pOH = -log10[OH-]
In this case, the weak acid is partially dissociated, so we will assume that the concentration of [OH-] is negligible. Therefore, the pOH of the solution is approximately 0.

3. Convert pOH to pH using the equation:
pH = 14 - pOH
pH = 14 - 0
pH = 14

So, the pH of the solution is 14.

B. To calculate the Ka of the acid:
1. Start with the dissociation equation of the weak acid:
HA ⇌ H+ + A-

2. Write the expression for Ka, which is the equilibrium constant for the dissociation of the weak acid:
Ka = [H+][A-] / [HA]

3. In the given problem, the concentration of the weak acid after dissociation is 0.0075 M, and since the weak acid dissociates partially, we can assume that the concentrations of [H+] and [A-] are equal to 0.0075 M. The concentration of the undissociated weak acid, [HA], can be calculated by subtracting the concentration of dissociated acid, 0.0075 M, from the initial concentration, 0.25 M:
[HA] = 0.25 M - 0.0075 M = 0.2425 M

4. Substitute the values in the expression for Ka:
Ka = (0.0075 M)(0.0075 M) / (0.2425 M) = 0.231 M

So, the Ka of the acid is approximately 0.231 M.

A. To calculate the pH of the solution, we need to determine the concentration of the hydronium ion (H3O+). Since the acid is weak, we can assume that the dissociation reaction is as follows:

HA ⇌ H+ + A-

Given that the acid is 3.0% dissociated, we can assume that the concentration of the hydronium ion is also 3.0% of the initial concentration.

Step 1: Calculate the concentration of H3O+ (c_H3O+):
c_H3O+ = 3.0% × [HA]

The concentration of HA can be determined using the given molarity (M) and volume (V) of the solution:
[HA] = M × V

Step 2: Calculate the pH using the concentration of H3O+:
pH = -log(c_H3O+)

B. To calculate the Ka of the acid, we need to use the relationship between Ka, the initial concentration of the acid (HA), and the concentration of the dissociated acid ([A-]).

The equilibrium constant expression for the dissociation of the weak acid is as follows:

Ka = [H+][A-] / [HA]

Given that the acid is 3.0% dissociated, we can assume that the concentration of [H+] is also 3.0% of the initial concentration of HA.

Step 1: Calculate the concentration of [H+] using the concentration of HA and the percent dissociation:
[H+] = 3.0% × [HA]

Step 2: Calculate the concentration of [A-] using the concentration of [H+] and the percent dissociation:
[A-] = 3.0% × [HA]

Step 3: Calculate the Ka using the concentrations of [H+], [A-], and [HA]:
Ka = ([H+][A-]) / [HA]