Fundamental Principles Of Stoichiometry, Buffer Solutions, And PH Calculations

Diluting Solutions

This problem sheet has calculations that involve the fundamental principles of solution stoichiometry, buffer solutions and pH. A variety of units are used in the questions and you need to check these carefully. Look out for SI prefixes such as µ (micro), m (milli) and k (kilo). Your answers should be written out in full with all intermediate steps shown and a clear explanation of your methodology. Ensure that all numerical quantities are shown with units and that your final answer is rounded to an appropriate number of significant figures. Where appropriate, assume a value of 1×10-14 for Kw, the ionic product of water. As each sheet is different, you MUST hand your sheet in with your answers.

Diluting solutions 1. You are required to make 1600 µL of 5 mM solution of a protein from a stock solution of concentration 25 mM. What volume of stock solution would you need to use? [units µL] 2. You are required to make 1000 mL of a protein solution containing 7 µg mL-1 from a stock solution containing 60 mg mL-1. What volume of stock solution would you need to use? [units mL] 3. What is the concentration of the final solution when 15 mL of a glucose solution containing 1.2 g L-1 of glucose is diluted by adding 190 mL of water? [units mmol L-1]

Hydrogen ions, hydroxide ions and pH

4. What is the hydroxide ion concentration in a solution if the hydronium ion concentration is 6.481×10-10 mol L-1?

5. What is the hydronium ion concentration in a solution if the hydroxide ion concentration is 8.426×10-4 μmol L-1?

6. What is [OH- ] in a solution where [H3O + ] is 2.872×10-5 mmol L-1?

7. What is [H3O + ] in a solution where [OH- ] is 3.694×10-8 mol L-1?

8. What is the pH of a solution of whose hydronium ion concentration is 1.675×10-6 mol L-1?

9. What is the pH of a solution of whose hydroxide ion concentration is 7.23×10-2 mmol L-1?

10. What is the pH of a solution of whose hydronium ion concentration is 6.652×10-1 μmol L-1?

11. What is the pH of a solution of whose hydroxide ion concentration is 3.384×10-6 mmol L-1?

12. Calculate the hydronium ion concentration of a solution of pH 3.18. [units mol L-1]

13. Calculate the hydroxide ion concentration of a solution of pH 2.99. [units mol L-1]

14. Calculate the [H3O + ] of a solution of pH -0.39. [units mol L-1]

15. Calculate the [OH- ] of a solution of pH 6.5. [units mol L-1]

Problems using pKa and the Henderson-Hasselbalch Equation

16. A weak acid has a pKa of 4.157. Calculate the pH of a solution in which the ratio of the concentrations of undissociated acid to acid anion ([HA]/[A- ]) equals 9.25.

17. A weak acid has a pKa of 4.597. Calculate the pH of a solution in which the ratio of the concentrations of acid anion to undissociated anion ([A- ]/[HA]) equals 4.108 (note the slight difference from the previous question).

18. A weak acid has a pKa of 5.299. If the solution pH is 5.723, what percentage of the acid is undissociated?

19. You need to prepare a buffer solution of pH 4.293 from 50 mL of 0.245 M solution of a weak acid whose pKa is 4.04. What volume of 0.269 M NaOH would you need to add? [units mL]

20. You need to prepare a buffer solution of pH 3.714 from 100 mL of 0.352 M solution of a sodium salt of a weak acid, Na+A - where the pKa of the weak acid HA is 3.79. What volume of 0.324 M HCl would you need to add? [units mL]

Diluting Solutions

Calculations of the Fundamental Principles of Stoichiometry, Buffer Solutions and pH.

Diluting Solutions

1. 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

1. 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 K_w 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.

K_w 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

K_w 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

K_w 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.

K_w 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

K_w 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= pK_a+ 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= pK_a+ log₁₀ ([A]/ [HA])

Hence, 5.587 + log (4.108)

pH= 5.211

 18. The pH – pK_a = log₁₀([A]/[HA])

5.299-5.723 = log₁₀([A]/[HA])

10^0.424 = [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 – pK_a = log₁₀([A]/[HA])

4.293- 4.04 = log₁₀([A]/[HA])

0.253 = log₁₀([A]/[HA])

10^0.253= 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 – pK_a = 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 [12^th April, 2018]

WinterChemistry. (2012). Acid and Bases Neutralisation. Website Online:

www.winterchemistry.com/wp-content/uploads/2012/01/PH-Notes-Ch.-2021.pdf   Retrieved [ 12^th April,2012].