Calculate the pH of 0.0115M acetic acid solution at 25 degrees celcius with Ka = 1.75*10^-5
To calculate the pH of an acetic acid solution, we need to use the dissociation constant (Ka) of acetic acid. The dissociation equation for acetic acid is as follows:
CH3COOH ↔ CH3COO- + H+
Given that the Ka value for acetic acid is 1.75 * 10^-5, we can set up an equation to solve for the concentration of H+ ions and, consequently, the pH.
First, we need to determine the concentration of CH3COOH that ionizes. Since the initial concentration of acetic acid is 0.0115 M, we can assume that the concentration of the dissociated CH3COO- ions is negligible compared to the initial concentration of acetic acid. Therefore, we can assume that the concentration of CH3COOH is approximately equal to the initial concentration.
Next, let's assume that the concentration of H+ ions is x M. Since the acid is weak, we can assume that the concentration of H+ ions is equal to the concentration of CH3COOH that has ionized.
Therefore, the mathematical expression for this step is:
[CH3COO-] = [H+] = x
Since acetic acid ionizes according to the dissociation equation, the concentration of CH3COOH that ionizes is equal to the initial concentration minus x, i.e. [CH3COOH] = 0.0115 - x.
Now, we can use the Ka expression to set up an equation:
Ka = [CH3COO-] * [H+] / [CH3COOH]
Inserting the previously mentioned values, we have:
1.75 * 10^-5 = x * x / (0.0115 - x)
Next, we solve this equation for x, the concentration of the H+ ions:
x^2 = 1.75 * 10^-5 * (0.0115 - x)
Simplifying, we get:
x^2 = 1.92 * 10^-7 - 1.75 * 10^-5x
Rearranging the equation:
x^2 + 1.75 * 10^-5x - 1.92 * 10^-7 = 0
This is a quadratic equation. By solving it, we find that x ≈ 0.004.
Now that we know the concentration of H+, we can calculate the pH using the formula:
pH = -log[H+]
Thus, the pH of the acetic acid solution with a concentration of 0.0115 M is pH = -log(0.004) ≈ 2.4.