Question

Calculate the pH of 100 mL of a buffer solution containing 0.05 M benzoic acid (C6H5CO2H; Ka = 6.4 x 10-5) and 0.05 M sodium benzoate (NaC6H5CO2).(Note: please report your pH value to two decimal places.) ***I got this part***

(f) Calculate the pH after adding 10 mL of 0.10 M NaOH to the solution in part (d).

Answer 1

so subtract the moles of acid neutralized by NaOH, and them to the moles of the conjugate base. Plug the values back into the Henderson-Hasselbalch equation.

Answer 2

sorry but not sure what to do still

Answer 3

how many moles of NaOH did you add?

How many moles of NaC6H5CO2 and C6H5CO2H did you have before you added NaOH?

Answer 4

.0001 moles NaOH?

.005 moles of others?

Answer 5

okay, so you neutralized some of the acid

new moles of acid=0.005-0.0001

those moles became the conjugate base of the buffer

new moles of base=0.005+0.0001

find new molarity (take the volume of NaOH into account)

plug back into formula

Answer 6

I got 12 and it says it's wrong. I am lost on this one.

Answer 7

\(pH=pKa+log(\dfrac{(\dfrac{0.005+0.0001}{0.11L})}{(\dfrac{0.005-0.0001}{0.11L})}\)

Answer 8

That gives me 4.28 which it counts wrong...

Answer 9

rounded wrong it was 4.29. thanks!

Answer 10

no problem !