lecture 3 sample problem answers

8/18/2019 Lecture 3 Sample Problem Answers

1/2

Biosc 1000

Sample Problem Answers Lecture 3

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pH =11, [H + ] = 1 !10

"11 M

K w = 1 ! 10"14

M 2 = [H

+ ][OH

"

]

1 ! 10"14

M 2 = [OH

"

] = 1 ! 10"3

M

1 ! 10"

11 M

3. Calculate the pH of solutions (made in distilled water) of

a) 0.1M HCl

0.1 M HCl is a strong acid and completely dissociates producing 1 X 10 –1

M of H + ,

which is a pH = 1

b) 300 ml of 0.25M sodium ascorbate plus 150 mL of 0.2 M HCl (The pKa of ascorbic acid

is 4.04)

Ascorbic acid is a weak acid (HA) & sodium ascorbate is its conjugate base (A-).The initial amount of ascorbate is 0.3L(0.25M) = 0.075 moles or 75 mmoles.

The number of moles of HCl added is 0.15L(0.2M) = 0.03 moles or 30 mmoles.

Since HCl is a strong acid, it reacts completely with the ascorbate, so total [HA] = 30

mmoles.

The amount of [A-] remaining is 75 mmol – 30 mmol = 45 mmol.

pH = 4.04 + log (45 mmol/30 mmol) = 4.04 + 0.18 = 4.22

Note: The values for mmol were used instead of M as usually used. This is valid because

the total volume is the same and would cancel out.

4. For a weak acid with a pKa of 6.4, what is the ratio of acid (HA) to base (A-) at pH 7.2?

pH = pKa + log ([A- ]/[HA])

=pKa – log ([HA]/[A- ])

log ([HA]/[A-])= pKa –pH

([HA]/[A-])= 10(pKa – pH)

= 10(6.4-7.2)

=10-0.8

= 0.16

So there are about 6 molecules of A- for every 1 of HA

8/18/2019 Lecture 3 Sample Problem Answers

2/2

Biosc 1000

5. At maximum buffering capacity, what is the ratio of a weak acid to its conjugate base?

A weak acid has its maximum buffering capacity at its pKa. Using the Henderson-

Hasselbalch equation, we find


If pH = pKa, then pH-pKa = 0 = log([A- ]/[HA]), so [A

- ]/[HA] must be 1

6. What volume (in mL) of 6 M acetic acid would have to be added to 500mL of a solution of

0.20M sodium acetate in order to achieve a pH = 5.0? The pKa of acetic acid is 4.75.

What are we trying to determine? The amount of HA to add to a 500 mL solution


To get a final pH of 5, what is the ratio of [A- ]/[HA]?

log ([A- ]/[HA]) = pH – pKa = 0.25

[A- ]/[HA] = 1.77

start with 0.1 mol acetate [A-]:

(0.2M) x 0.5 L = 0.1 mol

For [A-]/[HA], we can assume that all acid added stays associated. Therefore

all acid added stays as HA because whatever become dissociated will

reassociate; the amount of acid is so much less than the amount of

acetate.

Therefore:

0.1mol /[HA] = 1.77

[HA]= 0.1/1.77 = 0.056 mol of acid added

How much 6 M acetate to add?

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lecture 3 sample problem answers

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