NOVEMBER 2009
CHEMISTRY 265
T. SMITH-PALMER
[2 ] 1. Why are buffers important – what is their “claim to fame”?
Keep pH constant.
[2]
Explain in a few sentences how they are able to do this.
They contain acid and its conjugate base.
If acid is added it reacts with the base.
If base is added it reacts with the acid.
[1]
What is meant by buffer capacity?
The concentration or amount of acid and base present.
The more present—the more there is to react with large amounts of acid or base added.
[2]
Why are phosphate buffers very common?
For phosphoric acid, pKa1 = 2.148
pKa2 = 7.198
pKa3 = 12.375
They can be used to make buffer.
The middle one is around pH 7 - so it’s useful in biology.
[2]
If you had a solution that contained 0.5 moles of acetic acid and 0.001 moles of its conjugate
base, (pKa for acetic acid = 4.756), would this constitute a buffer. Explain your answer.
pH = 4.8 + log
.001
.5
The ratio << 10 so not a buffer.
1
2.(a) We have prepared a solution of leucine (an amino acid with the formula
(CH3)2CHCH2CH(NH2)COOH) in water.
[2]
What are the 3 forms of the amino acid, written in terms of ‘H’ and ‘L’ ?
H2L+, HL, L-
[1]
Which form is leucine?
HL
[2]
Estimate the pH of the solution formed when the leucine was dissolved.
Ka1 = 4.7 x 10-3
Ka2 = 1.80 x 10-10
HL Intermediate form
[ H  ]  4.7  10 3  1.80  10 10
 9.19  10 7 M
pH = 6.04
[2]
(b) If you prepared a solution of glycine hydrochloride, with a formality of 0.10 M, what
would be the concentration of the L- form?
Ka1 = 4.47 x 10 -3
Ka2 = 1.67 x 10-10
H2L H+ + HL[H+] = [HL-]
[ H  ][ L ]
K a2 
[ HL ]
[ L ]  1.67  10 10 M
 1.67  10 10 M
2
[4]
(c) Calculate the pH of 0.010 F monosodium malonate (malonic acid =propanedioic acid)
K a 2  1.42  10 3
K a1  2.01  10 6
[H  ] 
K a1 K a 2 F  K a1 K w
K a1  F
(1.42  10 3 )( 2.01  10 6 )(0.0100)

0.00142  0.01
[H  ]  4.996  10 5

K w  K a 2 F
0.0285  10 9
 24.96  10 10
0.01142
 5.00  10 5 M
[3]
(d) Calculate the pH of a solution made by mixing 0.03 moles HNO3 and 0.04 moles NaOH,
with a final volume of 100 mL.
0.01 mole
 0.1 M OH
0.1L
pOH = 1.0
pH = 13.0
3
[3]
(e) Calculate the pH of a solution made by mixing 0.05 moles HCl and 0.04 moles of
sodium propanoate in a final volume of 100 mL.
Ka for propanoic acid = 1.34  10-5
0.01 moles H 
0.100 L
excess
pH  1.0
[3]
(f) Calculate the pH of a solution made by mixing 0.020 moles of HCl with 0.100 moles of
monosodium
malonate.
pKa1 = 2.847, pKa2 = 5.696
0.02 moles of HCl reacts with the NaHM to form 0.02 moles H2M, leaving 0.08 moles HM-
0.08
0.02
 2.847  log 4
pH  pK a1  log
 2.847  0.602
 3.45
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3. An aliquot of 25 mLs of a solution of disodium oxalate was titrated with 0.0500 M HCl.
The second equivalence point was detected after the addition of 40.00 mLs of hydrochloric
acid.
For oxalic acid, Ka1 = 5.4 x 10-2
Ka2 = 5.42 x 10-5
[2] What was the initial concentration of the disodium oxalate?
Ox2- + 2H+  H2O
Moles H+ = 0.05 mmol/mL * 40 mL = 2.00 mmol
O 2x  2H   H 2 0
mmol
 40 mL  2.00 mmol
mL
2.00mmol
 moles O 2x 
 1.00 mmol
2
1.00 mmol
[O 2x ] 
 0.0400 M
25 mL
moles H   0.05
So moles Ox2- = 2.00 mmol/2 = 1.00
mmol
[Ox2- = 1.00 mmol/25 mL = 0.040 M
[3] What was the initial pH before any base was added?
O x2  H 2 O  HO x  OH 
K b1 
Kw
K a2
1.01  10 14
 1.863  10 10
5
5.42  10
[ HO x ][OH  ]

assume [OH  ]  F , let [OH  ]  x
F  [OH ]
K b1 
x2
 1.863  10 10
0.040
 [OH  ]  2.73  10 6
K b1 
pOH  5.564
pH  8.436
5
[4] What was the pH at the second endpoint?
H 2 O x  H 2 O  H 3 O   HO x
K a1
[H 3 O  ][ HO x ]

 5.4  10  2

F  [H 3 O ]
5.4  10  2 
x2
0.01538  x
mod erates well  needs quadratic equation
F
25
 0.04  0.01538
65
x 2  8.31  10  4  5.4  10  2 x
0  x 2  5.4  10  2 x  8.31  10  4
x
 b  b 2  4ac
2a

 5.4  10  2  0.00292  0.00332
2

 0.054  0.0790
 0.0125  [ x ]  [H  ]
2
pH  1.90
[3] 4. Write the charge balance equation for a solution that is 0.02 F in disodium phosphate.
[ Na  ]  [H  ]  [OH  ]  [H 2 PO 4 ]  2[HPO 24 ]  3[PO 34 ]
[2]
Fill in the following mass balance equation for a solution that is 0.02 F in disodium
phosphate
[Na+] = 0.04 M = 2( [H 3 PO 4 ]  [H 2 PO 4 ]  [HPO 24 ]  [PO 34 ] )
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[3] 5. If you have a mix of metal ions in solution, but you just want to titrate one of them with
EDTA, is that possible? Always? What steps could facilitate this?
Yes, sometimes. Can titrate on in the presence of another if the Kf values are very different
and you control the pH.
6. For the titration of Sr2+ with EDTA, log Kf = 8.72.  Y 4  = 0.81 at pH 11
[2]
What is the conditional formation constant at pH 11?
K f  0.81  5.25  108
= 4.25 x 108
[2]
What does it mean when we write  Y 4  ?
It is the fraction of the EDTA present which has lost four protons.
[1]
Write out the full name of EDTA.
ethylene diamine tetra acetic acid
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