In: Chemistry
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.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.