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Calculate the pH of a solution that is 0.250 M in HClO4 and 0.125 M in CH3COOH (Ka = 1.8 x 10-5)

User WebDucer
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To calculate the pH of the solution, we need to compare the strengths of the two acids. HClO4 is a strong acid, which means it completely dissociates in water. CH3COOH is a weak acid, so we need to consider its dissociation using the Ka value.

Since HClO4 is a strong acid, it will fully dissociate, resulting in a concentration of H+ ions equal to 0.250 M.

For CH3COOH, we can use the Ka value to calculate the concentration of H+ ions. The equation for the dissociation of CH3COOH is:

CH3COOH ⇌ CH3COO- + H+

Using the Ka expression:

Ka = [CH3COO-][H+] / [CH3COOH]

We know the concentration of CH3COOH is 0.125 M, so let's assume the concentration of CH3COO- is x M. Since CH3COOH and CH3COO- are in a 1:1 ratio, the concentration of H+ ions will also be x M.

Substituting the values into the Ka expression:

1.8 x 10-5 = (x)(x) / 0.125

Solving for x:

x^2 = 0.125 * 1.8 x 10-5

x ≈ 0.003

So, the concentration of H+ ions from CH3COOH is approximately 0.003 M.

To find the total concentration of H+ ions, we sum the contributions from both acids:

Total [H+] = [H+] from HClO4 + [H+] from CH3COOH

Total [H+] = 0.250 M + 0.003 M

Total [H+] ≈ 0.253 M

Finally, we can calculate the pH using the formula:

pH = -log[H+]

pH ≈ -log(0.253)

pH ≈ 0.596

Therefore, the pH of the solution is approximately 0.596.
User Wangdq
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