Question

8) Sulfuric acid is a strong acid, and the pKa 2 of HSO - 4 is 1.92. What is the pH of a 0.100 M

NaHSO 4 solution?

9) What is the pH of 1.5 L of a vinegar that is 3% acetic acid by mass? (Ka= 1.8 x 10 -5 ).

Answer 1

Answer.8.

The NaHSO₄ salt undergoes complete ionization.

Thus, NaHSO₄ = Na + HSO₄^-

Even the HSO₄^- can undergo further ionization

HSO₄^- = H⁺ + SO₄^2-

Lets presume the concentration of H⁺ be (x); Given pKa₂ is 1.92; therefore Ka₂ = 10^-1.92 = 0.0120

HSO₄^- = H⁺ + SO₄^2- = 0.100 - x(HSO₄^-) , x(H⁺), x (SO₄^2-)

Let's solve the equation -

Ka₂ = x² / (0.100-x) = 0.120

(H⁺) = x

= {0.120 + (0.012² + 4 X 0.00120)^1/2} / 2 = 0.0292 M

Therefore, pH = - log 0.0292 = 1.54

Thus the pH of 0.100 M NaHSO₄ solution is 1.54.

Answer. 9.

First lets find the initial concentration of Acetic acid (Molecular weight = 60.05 g/mol) in Vinegar. Presuming the vinegar is a solution of Acetic acid in water, then the density is 1 g/mL. Therefore, if Vinegar contains 3% acetic acid by mass, then

1.5 L Vinegar X 1000 mL/1 L X 1 g/ X 3 g Acetic acid/ 100 g Vinegar X 1 mol Acetic acid/ 60.05 g Acetic acid X =

0.75 mol Acetic acid

Thus, for initial concentration, 0.75 mol/ 1.5 L = 0.50 M

CH₃COOH + H₂O = CH₃COO^- + H₃O⁺

Initial concentration of Acetic acid is 0.5 M and as the reaction comes to equilibrium the concentration of Acetic acid is 0.5 - x

The equilibrium constant is given by, Ka = [CH₃COO^-][H₃O⁺]/ [CH₃COOH]

Given Ka = 1.8 X 10^-5 ; Substituting the values in the equilibrium formula

1.8 X 10^-5 = x² / (0.5 - x)

By solving, you will get x = 0.0030 M = (H₃O⁺)

pH = - log (H₃O⁺) = - log (0.0030) = 2.5

Therefore the pH is 2.5.