given that HNo2 is a stron acid and k3(CH3Cooh = 1.75 * 10^-5 , 1.0mol CH3COOHwhat is the ph calcuate the ph
To calculate the pH, we need to consider the dissociation of acetic acid (CH3COOH) in water. Since HNO2 is a strong acid, it completely dissociates in water. However, acetic acid is a weak acid and only partially dissociates. The dissociation reaction is as follows:
CH3COOH + H2O ⇌ CH3COO- + H3O+
First, we need to calculate the concentration of CH3COOH (acetic acid) that has dissociated.
Since we have 1.0 mol of CH3COOH, and it is a weak acid that partially dissociates, we assume a small fraction (x) dissociates. Therefore, the concentration of CH3COOH that has dissociated is x mol/L.
Since acetic acid is a weak acid, we can use the approximation that [CH3COOH] ≈ [CH3COOH]initial.
Therefore, after dissociation, the concentrations of CH3COOH, CH3COO-, and H3O+ are:
[CH3COOH] = 1.0 - x mol/L
[CH3COO-] = x mol/L
[H3O+] = x mol/L
Using the equilibrium constant expression for the dissociation of acetic acid (Ka = [CH3COO-][H3O+]/[CH3COOH]), we have the following relationship:
Ka = [x][x]/(1.0 - x)
Given that Ka = 1.75 * 10^-5, we can solve for x using the quadratic equation:
1.75 * 10^-5 = x^2/(1.0 - x)
Simplifying this equation, we get:
1.75 * 10^-5 * (1.0 - x) = x^2
1.75 * 10^-5 - 1.75 * 10^-5x = x^2
x^2 + 1.75 * 10^-5x - 1.75 * 10^-5 = 0
Solving this quadratic equation using the quadratic formula, we find that x ≈ 0.00418 mol/L.
This means that approximately 0.00418 mol/L of acetic acid has dissociated into acetate ions (CH3COO-) and hydronium ions (H3O+).
To calculate the pH, we can use the equation:
pH = -log[H3O+]
Since [H3O+] ≈ x ≈ 0.00418 mol/L, we have:
pH = -log(0.00418) ≈ 2.38
Therefore, the pH is approximately 2.38.