Question

23) What is the pH of a solution prepared by mixing: 

0.20 moles of acetic acid 

0.40 moles of sodium acetate 

0.10 moles of sodium hydroxide 

in 1.0 L of solution 

a) 4.74 b) 4.14 5.34 d) 13.00 e) None of the above 

24) Barbituric acid (Ka = 9.8 X 10^-5) is titrated with 0.200 M NaOH. What is the initial pH of 20.0 mL of 0.100 barbituric acid

a) 2.50 b) 4.01 c) 8.35 d) 7.00 e) None of the above 

25) What is the pH in the titration above after 5.00 mL of 0.200 NaOH are added to the barbituric acid 

a) 2.50 b) 4.01 c) 8.35 d) 7.00 e) None of the above

Answer 1

Answer 2

24. The pH of weak barbituric acid is given pH = 1/2(pKa+logC) = 1/2[-log(9.8*10^-5) -log(0.1)] = 2.50

25.                  Barbituric acid + NaOH ----------> sodium barbiturate + H2O   

   initial              2 mmoles          1 mmoles           -                             -

after the reaction   1 mmlole       o mmoles             1 mmole              1mmole

Now the mixture forms acidic buffer and its pH is given by pH = pKa + log(sodium barbiturate/Barbituric acid )

                              pKa= -log(Ka) =-log(9.8*10_5) = 4.008

pH = pKa + log(sodium barbiturate/Barbituric acid ) = 4.008+ log(1/1) = 4.008 4.01

Answer 3

Answer 4

23. (e) None of the above. Acetic acid is weak acid, sodium acetate is just a salt and

sodium hydroxide is a strong base.

24 (a) 2.50

25 (b) 4.01

Answer 5