Final answer:
To determine the pH of a 0.25 M potassium benzoate solution, calculate the Kb using the provided Ka and the ion-product constant, then determine [OH-] and subsequently pOH, and finally, calculate pH using the relation pH + pOH = 14.
Step-by-step explanation:
To determine the pH of a 0.25 M solution of potassium benzoate (KC7H5O2), we need to understand that potassium benzoate is the salt of a weak acid (benzoic acid, HC7H5O2) and a strong base (KOH), and will therefore form a basic solution when dissolved in water. The disassociation of potassium benzoate in water can be represented as:
KC7H5O2 (aq) → K+ (aq) + C7H5O2- (aq).
The benzoate ion (C7H5O2-) is the conjugate base of the weak acid benzoic acid and will react with water to produce OH- ions (hydroxide ions):
C7H5O2- (aq) + H2O (l) ⇌ HC7H5O2 (aq) + OH- (aq).
The given Ka value for benzoic acid is 6.3×10⁻⁵ which allows us to calculate Kb for the benzoate ion using the equation
Kw = Ka × Kb, where Kw is the ion-product constant for water at 25 °C, and it is equal to 1.0 × 10⁻ⁱ⁴.
Kb for benzoate ion can therefore be calculated, and then this can be used to find [OH-] concentration. Once [OH-] is known, pOH can be found with -log[OH-], and finally pH can be calculated using the relationship pH + pOH = 14.