The pKa value of CH3COOH is 4.74, the pH of equimolar solution of acetic acid and sodium acetate is:

Explanation

To determine the pH of an equimolar solution of acetic acid (CH3COOH) and sodium acetate (CH3COONa), we need to consider the acid-base equilibrium between acetic acid and its conjugate base acetate ion. The pKa value of acetic acid (CH3COOH) is given as 4.74. The pKa is a measure of the acidity of a compound and is defined as the negative logarithm of the acid dissociation constant (Ka). In this case, we have an equimolar solution of acetic acid and sodium acetate. Sodium acetate (CH3COONa) is the salt of the conjugate base acetate ion (CH3COO-) and a sodium cation (Na+). When the salt dissolves in water, it dissociates into acetate ions and sodium ions. In the equimolar solution, the concentration of acetic acid and acetate ions will be the same. Since acetic acid is a weak acid, it will partially dissociate into acetate ions and release H+ ions. The Henderson-Hasselbalch equation relates the pH of a solution to the pKa and the ratio of the concentrations of the conjugate acid and base: pH = pKa + log([A-]/[HA]) In this case, [A-] and [HA] represent the concentrations of the acetate ion and acetic acid, respectively. Since we have an equimolar solution, the concentration of acetate ions ([A-]) will be equal to the concentration of acetic acid ([HA]). pH = pKa + log(1) pH = pKa Therefore, the pH of the equimolar solution of acetic acid and sodium acetate will be equal to the pKa value of acetic acid, which is 4.74. Using Henderson's Equation: pH = pKa + log ([Salt]/[Acid]). where; Acid= Acetic Acid Salt= Sodium acetate [Salt]/[Acid]= 1 (as equimolar), hence log ([Salt]/[Acid])= 0 Therefore, pH= pKa= 4.74