Question

A.) What is the pH of a 500mL sample of 0.450M acetic acid solution, given a Ka=1.8x10^-5?

B.) What is the pH after 20.000ml of 0.8M NaOH are added?

C.) What is the pH after 140.625mL of 0.8M NaOH are added?

Answer 1

A)

pH =

Ka = [H+][A-]/[HA]

1.8*10^-5 = x*x/(0.45-x)

x = [H+] = 0.002837

pH = -log(H) = -log(0.002837) = 2.54714

B)

after V = 20 ml of M = 0.8 M is added

there is formation of buffer

[A-] = (0 + 0.8*20) = 16

[HA] = 0.45*500 - 0.8*20 = 209

pH = pKa + log(a-/HA

ph = 4.75 + log(16/209) = 3.6339

c)

after 140.625 ml

after V = 20 ml of M = 0.8 M is added

there is formation of buffer

[A-] = (0 + 0.8*140.625 ) = 112.5

[HA] = 0.45*500 - 0.8*140.625 = 112.5

pH = pKa + log(a-/HA) = pKa + log(1)

ph = 4.75