Answer to Question #121207 in Biochemistry for Nicolette DiGregorio

Question

1) You create 200 mL of a 0.35M buffer with a pH of 4.00. You add 8.0 mL of a 0.75M NaOH solution to the buffer you made, what is the new pH?

2) You create 200 mL of a 0.35M buffer with a pH of 4.00. You add 8.0 mL of a 0.75M HCl solution to the buffer you made, what is the new pH?

3) You create 50 mL of a 1.0M buffer with a pH of 4.7. You add 2 mL of a .45M NaOH solution to the buffer you made, what is the new buffer [Base]?
4) You create 150 mL of a 0.88M buffer with a pH of 3.3. You add 20.0 mL of a 0.35M HCl solution to the buffer you made, what is the new Buffer [Base]?

5) You create 50 mL of a 1.0M buffer with a pH of 4.7. You add 2 mL of a .45M NaOH solution to the buffer you made, what is the new buffer [Acid]?

Answer 1

$pH=pka+log(A)/(HA)$

1)convert the ml into litres.we will get 0.002l of buffer and 0.008l.The difference will be 0.192

pH=4.0+log(0.008/0.192)

pH=4.0- 1.380=2.62

2)pH=4.0+log(0.008/0.192)

=2.62

3) pH=4.7+log(0.002/0.048)

=4.7+1.38=6.08

4) pH=3.3+log (0.15/0.13)

=3.3+0.81=4.11

5) pH=4.7+log (0.002/0.048)

=4.7-1.38=3.32