Diluting Solutions
Calculations of the Fundamental Principles of Stoichiometry, Buffer Solutions and pH.
Diluting Solutions
- Calculate the volume of the stock solution needed to make 1600 micro litres solution of 5mM.
Dilution changes the concentration of the solution but does not change the amount of solute in the solution. Moreover, the product of the stock solution and its volume is equivalent to the product of the new concentration and its volume.
Concentration of stock solution Volume of stock = Concentration of new solution × Volume of the new solution
5mM × 1600 μL = 25 Mm × V
V =
V = 320 μL
- Calculate the volume of the stock solution needed to make 1000ml of protein of 7μg/ml.
The calculations should be in similar units. The concentration of the stock solution is 60mg/ ml hence, it is converted to μg/ml.
1mg = 1000μg
60mg =?
× 1000μg
60000μg
7μg × 1000 ml = 60000μg × V
2. Calculate the volume of the stock solution needed to make 1000ml of protein of 7μg/ml.
The calculations should be in similar units. The concentration of the stock solution is 60mg/ ml hence, it is converted to μg/ml.
1mg = 1000μg
60mg =?
× 1000μg
60000μg
7μg × 1000 ml = 60000μg × V
V = 0.116666666ml
V = 0.1167ml
3. Calculate the molarity of the final concentration after dilution in mmol/litre
The amount of solute in the solution remains constant after dilution.
Therefore, the amount of glucose in the 15ml solution is calculated first.
1.2g = 1000ml
? = 15ml
= 0.018g of glucose
190 ml of water is added to the 15 ml solution, making the new volume of solution to be 205ml.
0.018g of glucose are the ones still in the 205ml hence,
205ml = 0.018g
1000ml =?
= 0.087804878g/l
1 mole of glucose is equivalent to 180. 1559 g
180.1559g = 1 mole
0.087804878g =?
= 0.0004873827504moles/litre
However, 1 mole = 1000millimoles
0.0004873827504 moles =?
0.0004873827504/1 × 1000millimoles
= 0.48738275 millimoles/litre
= 4.874 × 10-1 mmoles/litre
Hydrogen ions, hydroxide ions and pH Calculations
4. Calculate the concentration of the hydroxide ions in the question
Water is not neutral because it does not have any hydronium ion or hydroxide ions absent but because these ions exist in water in equal concentration (10-14). This makes the Kw equivalent to 1.0 × 10-14 = [H3O+] [OH–]
Thus,
[OH–] = 1.0 × 10-14 / 6.481 × 10-10
= 1.542971764 × 10-5 mol/litre
= 1.543 × 10-5 mol/litre
5. Calculate the concentration of the hydronium ion.
Kw equivalent to 1.0 × 10-14 = [H3O+] [OH–]
Hydrogen ions, hydroxide ions and pH Calculations
[H3O+] = 1.0 × 10-14/ [OH–]
= 1.0 × 10-14/ 8.426 × 10-4 μmol/litre
= 1.186802753 × 10-11 μmol/litre
= 1.187 × 10-11 μmol/litre
6. Calculate the concentration of the hydroxide ions in the solution
Kw equivalent to 1.0 × 10-14 = [H3O+] [OH–]
Thus,
[OH–] = 1.0 × 10-14 /2.872 × 10-5 mmol/litre
=3.48189415 × 10-10mmol/litre
= 3.482 × 10-10mmol/litre
7. Calculate the hydronium ion concentration in the solution
Kw equivalent to 1.0 × 10-14 = [H3O+] [OH–]
[H3O+] = 1.0 × 10-14/ [OH–]
= 1.0 × 10-14/ 3.694 × 10-8mol/litre
= 2.707092583 × 10-7mol/litre
= 2.707 × 10-7mol/litre
8. Calculate the pH of the solution
pH is a scale of measure from 0-14 that indicates the number of hydrogen ions in a solution.
pH + p OH = 14
p OH = – log[OH–]
Thus, the concentration of the hydroxide ions is first calculated.
Kw equivalent to 1.0 × 10-14 = [H3O+] [OH–]
Thus,
= 1.0 × 10-14/ 1.675 × 10-6 mol/litre
= 5.9070149254 × 10-9 mol/litre
The p OH = -log [OH–]
Hence, – log (5.9070149254 × 10-9)
= 8.224014811
The pH + p OH = 14
Hence, p H = 14 – p OH
= 14 – 8.224014811
= 5.775985189
= 5.776
9. Calculate the pH from the given concentration of the hydroxide ions
The pH is worked out using the SI units of mols/litre
The p OH = -log [OH–]
= – log (7.23 × 10-5mol/l)
=4.140861703
Thus, the pH is 14- p OH
= 14 – 4.140861703
= 9.859
10. Calculate the pH given the hydronium ion concentration
The pH is worked out using the SI units of mols/litre
Kw equivalent to 1.0 × 10-14 = [H3O+] [OH–]
Thus,
= 1.0 × 10-14/ 6.652 × 10-1μmol/l
= 1.503307276 ×10-20mol/l
The p OH = – log [OH]
= -log (1.503307276 ×10-20mol/l)
=19.82295224
pH = 14 – p OH
= 14 – 13.82295224
= -5.823
= 0.1770
11. Calculate the p H given the hydroxide concentration
The p OH = – log[OH]
= – log ( 3.386 × 10-6)
= 5.470313046
The pH = 14 – p OH
= 14 – 5.470313046
= 8.529686954
= 8.530
12. The p OH = 14 – p H
= 14 – 3.18
= 10.82
10-p OH = [OH–]
= 10-10.82
[OH–] = 1.513561248 × 10-12mol/l
The hydronium concentration = 1.0 × 10-14/ 1.513561248 × 10-12mol/l
= 6.607 × 10-4mol/l
13. The p OH = 14 – 2.99
= 14.39
10-p OH = [OH–]
= 10 -14.39
= 9.772 × 10-12mol/l
14. The p OH = 14 – pH
= 14 + 0.39
= 14.39
10-p OH = [OH–]
10-14.39
= 4.073802778 × 10-15mol/l
The hydronium concentration = 1.0 × 10-14/ 4.073802778× 10-15
= 2.455mol/l
15. Calculate the [OH]
The p OH = 14 – p H
= 14- 6.5
= 7.5
10-p OH = [OH–]
10-7.5
= 3.162 mol/l
Hasselbach Calculations
16. Calculate the p H of the solution.
The pH= pKa+ log10 ([A]/ [HA])
[HA]/[A] = 9.25 Hence, [A]/[HA] = 1/9.25
The p H = 4.157 + log10 (1/9.25)
= 4.157 – 0.966141732
pH= 3.191
17. The pH of the solution is given by pH= pKa+ log10 ([A]/ [HA])
Hence, 5.587 + log (4.108)
pH= 5.211
18. The pH – pKa = log10([A]/[HA])
5.299-5.723 = log10([A]/[HA])
100.424 = [A]/[HA]
= 2.655 = 2655/1000
Percentage of the undissociated acid is therefore equals to 1000/3655 ×100%
=27.36%
19. Calculate the volume of the NaOH needed
The pH – pKa = log10([A]/[HA])
4.293- 4.04 = log10([A]/[HA])
0.253 = log10([A]/[HA])
100.253= 1.791
Concentration of weak acid × Volume of the weak acid = Concentration of the Conjugate acid × Volume of the conjugate acid
50ml × 0.254M = 1791M × V
V = 50ml × 0.254M/1791M
V= 7.093ml
20. Calculate the volume needed
The pH – pKa = log10([A]/[HA])
3.714-3.79 = log10([A]/[HA])
-0.076 = log10([A]/[HA])
10-0.076= 0.8395
Concentration of weak acid × Volume of the weak acid = Concentration of the Conjugate acid × Volume of the conjugate acid
100ml × 0.352M = 0.8395M × V
V = 100ml × 0.352M/ 0.8395M
V = 41.93 ml
References
Clark, J. (2016). Buffer Solutions. Physical Chemistry, Website Online:
www.chemguide.co.uk/physical/acidbaseeqia/buffers.html retrieved [12th April, 2018]
WinterChemistry. (2012). Acid and Bases Neutralisation. Website Online:
www.winterchemistry.com/wp-content/uploads/2012/01/PH-Notes-Ch.-2021.pdf Retrieved [ 12th April,2012].
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