Consider the titration of 100.0 mL of 0.200 M acetic acid ( K a = 1.810 5 ) by 0
ID: 537136 • Letter: C
Question
Consider the titration of 100.0 mL of 0.200 M acetic acid (Ka = 1.8105) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. Enter all answers to two decimal places.
(a) 0.0 mL
(b) 50.0 mL
(c) 100.0 mL
(d) 150.0 mL
(e) 200.0 mL
(f) 250.0 mL
AND
A buffer solution with a volume of 0.0230 L consists of 0.67 M iodous acid (HIO2), a weak acid, plus 0.67 M lithium iodite (LiIO2). The acid dissociation constant of iodous acid, Ka, is 3.2105. Determine the pH of the buffer solution after the addition of 0.0031 mol sodium hydroxide(NaOH), a strong base. (Assume no change in solution volume.)
Explanation / Answer
(1)
Moles of acid present initially = 0.1*0.2 = 0.02 moles
Moles of salt present initially = 0 moles
(a) 0 mL:
The acid dissociates as:
HA -----> H+ + A-
Initial 0.2 0 0
Eqb 0.2-x x x
Ka = x2/(0.2-x) = 1.8*10-5
Solving, we get:
x = [H+] = 0.0019 M
So, pH = -log([H+]) = 2.72
(b) 50 mL
Moles of base added = 0.05*0.1 = 0.005 moles
After addition of these many moles of base, moles of acid decrease and moles of salt increase by this amount.
So,
Moles of acid finally = 0.02-0.005 = 0.015
Moles of salt finally = 0 + 0.005 = 0.005
Using Henderson Hasselbach equation:
pH = pKa + log(moles of salt/moles of acid)
So,
pH = 4.74 + log(0.005/0.015) = 4.26
(c) 100 mL
Moles of base added = 0.1*0.1 = 0.01 moles
After addition of these many moles of base, moles of acid decrease and moles of salt increase by this amount.
So,
Moles of acid finally = 0.02-0.01 = 0.01
Moles of salt finally = 0 + 0.01 = 0.01
Using Henderson Hasselbach equation:
pH = pKa + log(moles of salt/moles of acid)
So,
pH = 4.74 + log(0.01/0.01) = 4.74
(d) 150 mL
Moles of base added = 0.15*0.1 = 0.015 moles
After addition of these many moles of base, moles of acid decrease and moles of salt increase by this amount.
So,
Moles of acid finally = 0.02-0.015 = 0.005
Moles of salt finally = 0 + 0.015 = 0.015
Using Henderson Hasselbach equation:
pH = pKa + log(moles of salt/moles of acid)
So,
pH = 4.74 + log(0.015/0.005) = 5.21
(2)
Moles of acid present initially = Molarity*Volume = 0.67*0.0230 = 0.01541
Moles of salt present initially = 0.67*0.0230 = 0.01541
Moles of base added = 0.0031
After addition of these many moles of base, moles of acid decrease and moles of salt increase by this amount.
So,
Moles of acid finally = 0.01541-0.0031 = 0.01231
Moles of salt finally = 0.01541+0.0031 = 0.01851
Using Henderson Hasselbach equation:
pH = pKa + log(moles of salt/moles of acid)
So,
pH = 4.49 + log(0.01851/0.01231) = 4.67