Consider the titration of 100.0 mL of 0.82 M H 3 A by 0.82 M KOH for the next th
ID: 877492 • Letter: C
Question
Consider the titration of 100.0 mL of 0.82 M H3A by 0.82 M KOH for the next three questions. The triprotic acid has Ka1 = 1.0 x 10-5, Ka2 = 1.0 x 10-8, and an unknown value for Ka3.
1) Calculate the pH after 100.0 mL of KOH has been added.
pH =
2) Calculate the pH after 150.0 mL of KOH has been added.
pH =
This is one of the three equivalence points in the titration. At the first two equivalence points, an amphoteric substance is the only major species present, with acidic/basic properties. To calculate the pH when an amphoteric substance is the only major species, we use the formula pH = (pKa1 + pKa2)/2 or pH = (pKa2 + pKa3)/2, depending on which amphoteric substance is present.Explanation / Answer
1). Ka1 = 1.0X10-5
pKa1 = - logKa1
= - log(1.0X10-5 )
= 5
Ka2 = 1.0X10-8
pKa2 = - log(1.0X10-8 )
= 8
pH = (pKa1 + pKa2)/2
= (5 + 8) /2
= 13/2
= 6.5
2). Volume of KOH added = 100mL + 150 mL = 250 mL
= 0.250 L
Moles of OH- = 0.82*0.250
= 0.205
Moles of OH- in excess = 0.205 - 0.082
= 0.123 moles
Total volume = 100 + 250 = 350 mL
= 0.350L
[OH-] = 0.123/0.350 = 0.351
pOH = - log[OH-]
= - log 0.351
= 0.45
pH = 14 - pOH
= 14 - 0.45
= 13.55