Question

8. Calculate the final pH of a solution made by the addition of 10 mL of a 0.5 M NaOH solution to 500 mL of a weak acid, HA. The concentration of the conjugate acid is 0.2M, the pH is 5.00 and the pKa = 5.00. Neglect the change in volume

Answer 1

g. 6.10
h. 5.04
i. 7.00
j. 5.99
k. 6.91

Answer 2

Use the Henderson-Hasselbalch equation

Answer 3

i have the problem worked out, but I have trouble with the beginning part of it. this is what they put
10 mL 0.5 M =0.005 M 0.005/0.5=0.01 M

5.00=5.00+ log [B]/[A] [B]=0.2 =[A]

Answer 4

10 mL * 0.5 M = 0.005 moles

0.005 mole/0.51 L=0.009803 M \(\approx\) 0.01 M

Answer 5

is that what was confusing you?

Answer 6

o ok i c. they converted the 10mL to L

Answer 7

yeah, then found moles. They incorrectly used M for moles

Answer 8

can u explain this part?5.00=5.00+ log [B]/[A] [B]=0.2 =[A]
pH =5.00 + log[0.2+0.01]/[0.2-0.01]=5.04

Answer 9

"The concentration of the conjugate acid is 0.2M" means that [B]=0.2 M

"the pH is 5.00 and the pKa = 5.00", when the pH=pKa is the pH where half of the species is present in 1 form and half in the other.

so [B]=[BH]=0.2 M


Then instead of using moles, they used molarity for the acid and base. They added base, so some of the acid was deprotonated (subtracted the molarity of the base NaOH form the acid) and added the same value to the conjugate base.

\(pH =5.00 + log\dfrac{[0.2+\color{red}{0.01}]}{[0.2-\color{red}{0.01}]}=5.04\)

Answer 10

ok thx i understand now

Answer 11

good stuff ! no probs