179k views
2 votes
What is the ph of a 0.183 m aqueous solution of nach3coo? ka (ch3cooh) = 1.8x10-5

1 Answer

1 vote

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.

User Greggz
by
7.6k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.