Final answer:
The pH of a 0.183 M aqueous solution of NaCH3COO can be calculated by first determining the hydrolysis constant (Kb) for the acetate ion, and then using the equilibrium expression for hydrolysis to find the OH- concentration. This can then be used to find the pOH, and finally, the pH is determined by subtracting pOH from 14.
Step-by-step explanation:
To calculate the pH of a 0.183 M aqueous solution of NaCH3COO, we need to understand that sodium acetate (NaCH3COO) is the salt of a weak acid (acetic acid, CH3COOH) and a strong base (NaOH). When NaCH3COO dissolves in water, it forms sodium ions (Na+) and acetate ions (CH3COO−). The Na+ ions will not affect the pH of the solution, as they are inert. However, the acetate ion will react with the water to produce acetic acid and hydroxide ions (OH−), which will increase the pH of the solution due to the basic nature of OH−.
To determine the pH, we would use the following equilibrium expression associated with the hydrolysis of the acetate ion:
CH3COO−(aq) + H2O(l) ⇌ CH3COOH(aq) + OH−(aq)
The concentration of acetate ions is approximately the same as the initial concentration of NaCH3COO. As we have the Ka value of acetic acid (1.8 × 10−5), we can calculate the hydrolysis constant (Kb) for the acetate ion using the relationship Kw = Ka × Kb where Kw is the ion product of water (1.0 × 10−14 at 25 °C).
Once we have Kb, we can use the equilibrium expression for the hydrolysis of acetate ions to find [OH−] and then calculate the pOH of the solution. The pH is then found by subtracting the pOH value from 14.