Question

In: Chemistry

8) Sulfuric acid is a strong acid, and the pKa 2 of HSO - 4 is...

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 ).

Solutions

Expert Solution

Answer.8.

The NaHSO4 salt undergoes complete ionization.

Thus, NaHSO4 = Na + HSO4-

Even the HSO4- can undergo further ionization

HSO4- = H+ + SO42-

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

HSO4- = H+ + SO42- = 0.100 - x(HSO4-) , x(H+), x (SO42-)

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.


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