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

The ratio < 10 so not a buffer.

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

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

H2Là H+ + HL-

[H+] = [HL-]

[4] (c) Calculate the pH of 0.010 F monosodium malonate (malonic acid =propanedioic acid)


[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.

pOH = 1.0

pH = 13.0

[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

[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-

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+ D H2O

Moles H+ = 0.05 mmol/mL * 40 mL = 2.00 mmol

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?

[4] What was the pH at the second endpoint?

[3] 4. Write the charge balance equation for a solution that is 0.02 F in disodium phosphate.

[2] Fill in the following mass balance equation for a solution that is 0.02 F in disodium phosphate

[Na+] = 0.04 M = 2()

[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. = 0.81 at pH 11

[2] What is the conditional formation constant at pH 11?

= 4.25 x 108

[2] What does it mean when we write ?

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

1