Question

A buffer solution is made that is 0.353 M in CH3COOH and 0.353 M in CH3COO- . (1) If Ka for CH3COOH is 1.80×10-5 , what is the pH of the buffer solution? (2) Write the net ionic equation for the reaction that occurs when 0.090 mol HI is added to 1.00 L of the buffer solution. Use H3O+ instead of H+ .

Answer 1

Answer:

(1) pH of the Buffer solution = 4.82

(2) pH of Buffer solution after addition of HI = 4.6

Explanation:

(1) A buffer solution consisting a mixture of a weak acid and its conjugate base is termed as Acidic Buffer.

CH₃COOH is a weak acid and its conjugate base CH₃COO⁻ is strong.

Ka (acid dissociation constant) = 1.80 x 10⁻⁵

⇒pKa = -Log(1.80 x 10⁻⁵ )

⇒pKa = 4.82

Using Henderson equation

pH = pKa + Log$\frac{[Salt]}{[Acid]}$

pH = 4.82 + Log$\frac{0.353}{0.353}$

pH = 4.82

(2) HI is a strong acid and dissociate completely.

  HI ⇆ H⁺ + I⁻

Due to its complete ionization a common ion effect arrises .

Assume all acid react with acetate ion to produce acetic acid (CH₃COOH)

CH₃COO⁻ + H₃O⁺ ⇄ CH₃COOH + H₂O

[H₃O⁺] added =  1 lit 0.09 mole = 0.09 molar

[CH₃COOH] = 0.353 + 0.09 = 0.443

[CH₃COO⁻]  = 0.353 - 0.09 = 0.263

Using Henderson equation

pH = pKa + Log$\frac{[Salt]}{[Acid]}$

pH = 4.82 + Log$\frac{0.263}{0.443}$

pH = 4.82 - .22 = 4.6