Answer: 2.78
Step-by-step explanation:
To calculate the pH of 0.16M CH3COOH, we need to use the dissociation constant (Ka) of the acid, which is given as 1.74 × 10^-5 mol dm^-3.
The dissociation of CH3COOH is as follows:
CH3COOH + H2O ↔ CH3COO- + H3O+
The equilibrium constant expression for the dissociation of CH3COOH is:
Ka = [CH3COO-][H3O+] / [CH3COOH]
We can assume that [H3O+] is equal to [CH3COO-] since the acid is weak and will not dissociate completely. Therefore, the equilibrium constant expression can be simplified as:
Ka = [H3O+]^2 / [CH3COOH]
[H3O+]^2 = Ka x [CH3COOH]
[H3O+]^2 = 1.74 × 10^-5 x 0.16
[H3O+]^2 = 2.784 × 10^-6
[H3O+] = √2.784 × 10^-6
[H3O+] = 0.00167 M
Therefore, the pH of the solution can be calculated as:
pH = -log[H3O+]
pH = -log(0.00167)
pH = 2.78
Therefore, the pH of the 0.16M CH3COOH solution is 2.78.