Question

2. Buffer Questions (Do 4 of 5 problems---6 pts each) a. Three different weak bases and their Kh values are listed below in a table. Which one would be best to form a buffer with a pH of 8.50? Explain why, and numerical calculations are expected as part of your answer to receive full credit. K = K,K) = 1.0 x 10-14 Weak Base K, Value KPK ammonia (1,8 x 10 hydrazine 1.3 x 10 pyridine 1.7 x 10 b. Write out the two chemical reactions that illustrate how a buffer (HA/A) functions: c. Describe three different chemical methods to make a H.CO/HCO, buffer use real chemicals and be specific with THIS example: d. Derive the Henderson-Hasselbach buffer equation using the weak acid equilibrium reaction and the K, expression for HA (aq) e. Three different weak acids and their K, values are listed below in a table. Which one would be best to form a buffer with a pH of 3.93? Explain why, and numerical calculations are expected as part of your answer to receive full credit. Weak Acid K, ValuepK, Acetic acid 1.8 x 103 Benzoic acid 6.5 x 10 Formic Acid 1.7 x 104

please solve 4 of them, thank you.

Answer 1

a)Given

NH3 , Kb = 1.8x10-5 , hence pKb = 4.75

N2H4 , Kb = 1.3x10-6, hence pKb = 5.89

pyridine Kb = 1.7x10-9, hence pKb = 8.77

From Hendersen equation pH of buffer is calculated as

pOH = pKb + log [conjugate acid]/[base]

assuming [conjugate acid] = [base]

pOH = pKB and pH = 14-pOH .

So required pH = 8.50 , that is pOH = 14-8.5= 5.5

Thus the best base tht can give the pOH around 5.5 is hydrazine with pKb value 5.89.

b)The buffer is HA/A-

When an acid is added (H+)

A- + H+ --------------> HA

thus the H+ is taken by A-(conjugate base) and hence no/small change in pH

when base (OH-) is added

HA + OH- ------------> A- + H2O

The added OH- is absorbed by undissociated acid to give A- and undissociated water , thus no/little change in pH.

d)

HA <------------> A- + H+

As in buffer we have the salt A- as (NaA, a strong electrolyte) is also added , due to the common ion effect the dissociation of weak acid HA is further supppressed.

Thus [acid] = [HA] at equilibrium and [A-] = [salt/conjugate base]] at equilibrium.

HA <------------> A- + H+

Thus the acid dissociation constant Ka = [A-][H+] /[HA] can be written as

Ka = [conjugate base] [H+]/[acid]

and

[H+] = Ka [acid]/[conjugate base]

we know pKa = -log Ka and pH = -log [H+}

Thus

pH = -log Ka - log [acid]/[conjugate base]

= pka - log [acid]/[conjugate base]

pH = pKa + log [conjugate base] /[acid]

e)

acetic acid Ka = 1.8x10-5 and pKa = 4.75

benzoic acid Ka = 6.6x10-5 , thus pKa = 4.19

formic acid Ka = 1.7x10-4 and pKa = 3.77

Using Hendersen equation

pH = pKa + log [conjugate base]/[acid]

and assuming [conjugate base] = [acid]  

the acid with pKa closer to 3.93 can make the best buffer , that is formic acid with pka = 3.77