SCHOOL OF BIOLOGICAL SCIENCES
BS1080 and BS1090 Module 3
Study Session 2
UNITS, pH AND pkA Answers
1.
2.
3.
If a solution of NaC1 contains 1.5g litre-1 what weights of NaC1 do the following volumes
of this solution contain?
(a)
10 ml
....15 mg or 0.015g......
(b)
5 μl
...... 7.5 g or 7.5 x 10-6g...
(c)
1 nl
......1.5 ng or 1.5 x 10-9g...
Given a 2.5 M NaC1 solution:
(a)
10 μl contains ..........25 μmol
(b)
2 ml contains .......... 0.005 mol NaC1
(c)
1 ml contains ..........2.5 mmol NaC1
(d)
20 μl contains .......... 50,000 nmol NaC1
(a)
If the molecular mass of glucose is 180 what is the mass of:
(b)
4.
(a)
(i)
1 mmol glucose ......0.18g
(ii)
5 μmol glucose ..........0.9 mg
How would you prepare:
(i)
1 litre of a 0.1 M solution of glucose ...18 g in 1 litre.
(ii)
100 ml of a 1 mM solution of glucose ......18 mg in 100 ml
How would you convert a 5 mM glucose solution to a 10 μM solution?
Dilute the 5mM solution 500 times
e.g 1 ml glucose plus 499 ml water
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(b)
1 ml of a 100 mM solution of NaC1 was diluted to a final volume of 10 ml to
give solution A.( !0-fold dilution) (10 mM)
2 ml of solution A was diluted to 5 ml to give solution B. (2.5-fold dilution)
(4mM)
1 ml of solution B was diluted to 2 ml to give solution C. (2-fold dilution) (2mM)
What would be the concentration of solution C? (2 mM)
Alternative method : Overall dilution = 10 x 2.5 x 2 = 50
Therefore 100 mM diluted to 2 mM
5.
What is:
(a)
the ionic product of water at 25oC? Kw = 1 x 10-14
(b)
the pH of water at 25oC?
pH = 7.0
Can you relate (a) and (b)?
[H+] = 10-7 ; in H2O at pH 7 [H+] = [OH-]
6.
7.
Kw = [H+] x [OH-]
= 10-7 x 10-7 = 10-14
What is the pH of:
(reminder
pH
(a)
1 M HC1
= -Log10 [H+])
pH = 0
(b)
0.5 M HC1
pH = 0.3
(c)
2 mM H2SO4
(d)
0.05 M KOH
(e)
1 M KOH
(a)
If a solution of HC1 has a pH of 3.5 what is its molarity?
pH = 2.4 ( reminder: 4mM H+)
pH = 12.7
pH=14.0
Log10 1/[H+] = - 3.5; Take Antilog. = 3.2 x 10-4 M
(b)
If the pH of a solution of KOH is 11 what is its molarity?
pH + pOH = 14;
pOH + 11 = 14; therefore pOH = 3
Antilog – 3.0 = 1 x 10-3M
(c)
If 2ml of 0.5 M H2SO4 is added to 10ml of 0.1 M KOH what is the pH of
2
the resulting solution?
2 ml M H2SO4 = 2 x 10-3 mol [H+]
10 ml of 0.1 M KOH= 1 x 10-3 mol [OH-]
Therefore un-neutralized H+ = 1 x 10-3 mol [H+]
But this has now been diluted to 12 ml
1 x 10-3 mol [H+] in 12 ml = 0.083 M
pH = 1.079
8.
(a)
Calculate the pH of a solution containing 0.5 M CH3CO2H and
0.5 M CH3CO2Na. The pKa of CH3CO2H is 4.76
pH = pKa + log10 [Ac-]/[HAc]
= 4.76 + log10 1
pH = 4.76
(N.B at 50% dissociation : pH = pKa
(b)
How would you prepare one litre of a 0.1 M acetate buffer, pH 5.5?
(relative molecular mass of acetic acid = 60, density 1.049g litre-1;
relative molecular mass of sodium acetate = 82)
Apply Henderson Hasselbalch Equation
pH = pKa + log10
[Ac-]
HAc]
5.5 = 4.76 + log10
[Ac-]
HAc]
5.5 - 4.76 = log10
[Ac-]
= 0.74 (take Antilog)
[HAc]
[Ac-]
= 5.50
[HAc]
Therefore to make 1 litre of 0.1 M sodium acetate / acetic acid buffer pH 5.5 we need:
8.2 g of sodium acetate (Mol Wt 82 x molar concn required)
and for acetic acid:
0.1
M ÷ 5.50 (molar ratio required) = 0.0182 M
i.e 60 x 0.0182 (Mol Wt 60 x molar concn required)
= 1.092g of acetic acid
3
But it is difficult to weigh out acetic acid and much easier to measure the volume as follows:
1.092 g ÷ 1.0499 kg . litre –1 (Mass ÷ Density)
= 1.040 ml of acetic acid
9.
Given that pK1 for the ionization of the carboxyl group of an amino acid, RCH(NH2)CO2H is 2.3, calculate the ratio of the ionized to unionized forms at pH 4.0 using
the Henderson-Hasselbach equation.
Using chemical formulae, indicate the likely changes in the structure of this compound
when the pH is raised from 1.0 through 6.0 to 12.0.
pH = pKa + log10 [R-COO-]/[R-COOH]
4.0 = 2.3 + log10 [R-COO-]/[R-COOH]
1.7 = log10 [R-COO-]/[R-COOH]
Antilog 1.7 = 50.1
= [R-COO-]/[R-COOH]
At pH 1.0
R- CH-( CO2H) NH3+
{I}
At pH 6.0
R- CH-( CO2-) NH3+
Zwitterionic Form {II}
At pH 12.0
R- CH-( CO2-) NH2
{III}
At PK1
{I}
=
{II}
At PK2
{II}
=
{III}
DRD/study session 1. Oct 2004
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