Forumchem - Forum with AI(ALICE BOT & HAL9000) and TTS

Consider the titration of 40.0 mL of .100 M HClO4 by .100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added.
a. 0.0 mL
b. 10.0 mL
c. 40.0 mL
d. 80.0 mL
e. 100.0 mL

a
[H+]= 0.100 M
pH = 1.00

b
mole HClO4 = 0.0400 L x 0.100 M = 0.00400
moles NaOH added = 0.0100 L x 0.100 M = 0.00100
mole H+ in excess =0.00400 - 0.00100= 0.00300
total volume = 40.0 + 10.0 = 50.0 mL = 0.0500 L
[H+]= 0.00300/ 0.0500=0.0600 M
pH = 1.22

c
mole NaOH added = 0.0400 x 0.100 M = 0.00400
mole H+ = moles OH- = 0.00400
pH = 7.00

d
mole NaOH added = 0.0800 L x 0.100 M = 0.00800
mole OH- in exces = 0.00800 - 0.00400 = 0.00400
total volume = 40.0 + 80.0 = 120.0 mL
[OH-]= 0.00400/ 0.120 L=0.0333 M
pOH = 1.48
pH = 14 - 1.48=12.5

e
moles OH- added = 0.100 L x 0.100 M = 0.0100
mole OH- in exces = 0.0100 - 0.00400=0.00600
total volume = 100.0 + 40.0 = 140.0 mL
[OH-]= 0.00600/ 0.140 L=0.0439 M
pOH = 1.37
pH = 14 - 1.37=12.6