Please help me with my chemistry. Our exam is in the morning.?

Question

18. Calculate the pH of 100 ml of 0.100 M formic acid (HCHO2) Ka = 1.8 x 10-4 (10 points)
19. Calculate the pH of a solution containing 100 ml of 0.200 M formic acid (HCHO2) and 100 ml of 0.180 M sodium formate (NaCHO2) (10 points)

20. Calculate the pH of the solution in Question 20 if 100ml of 0.100 M NaOH are added. (15 points)

Thank you so much! That makes perfect sense!

Answer 1

18.
Ka = [H+][HCOO-] / [HCOOH]
NB because HCOOH is monprotic (releases only one H) then
H+ = HCOO- so substituting in

Ka = [H+][H+] / [HCOOH]
Ka = [H+]^2 / [HCOOH]
[H+]^2 = Ka x [HCOOH]
[H+]^2 = 1.8 x 10-4 x 0.1
[H+]^2 = 1.8 x 10^-5
[H+] = sq rt {1.8 x 10^-5}
[H+] = 4.24 x 10^-3

pH = -log(10)[H+]
ph = - log(10)[4.24 x 10^-3]
pH = - -2.37
pH = 2.37

19.
Ka = [H+]{HCOO-] / [HCOOH]
[H+] = Ka x [HCOOH] / [HCOO-]
The conc'n of methanoate ions is now 0.2 + 0.18 = 0.38
[H+] = 1.8 x 10^-4 x 0.2 / 0.38
[H+] = 9.47 x 10^-5
pH = - log (10)[9.47 x 10^-5]
pH = - -4.02
pH = 4.02

20.
If 100 ml of 0.1M NaOH is added to the buffer, then there are 0.01 moles OH- ions. This will remove 0.01 H+ ions to water.
So 9.47 x 10^-5 - 0.01 = 9.91 x 10^-3 of OH- remains.

Kw = [H+][OH-] / [H2O] = 10^-14
[H+] = 10^-14 / 9.91 x 10^-3
[H+] = 1.01x 10^-12
pH = - log (10)[1.01 x 10^-12]
pH = - -12.0
pH = 12.0