Basics of Medical Chemistry Course

5. lecture. Consultation for solution of calculation problems – preparation for the 1st Midterm

1. Concentrations2. Titration problems

3. Ionization, pK, degree of dissociation4. pH of strong and weak acids

5. Buffers6. Solubility product, solubility

Andras HrabakDepartment of Medical Chemistry, Molecular Biology and

Pathobiochemistry

6 October 2009

1.1.4.7. What is the molarity of a 28 w / w% KOH solution (d = 1.27 g / cm3) ?

Mass/volume percent = w/w % * d = 35.56 w/v %

Therefore 100 ml solution contains 35.56 g KOH 1000 ml solution contains 355.6 g KOH Molar mass of KOH is 56.

355.6 / 56 = 6.35 M

2.1.2.9. Calculate the concentration of an unknown sulfuric acid solution in w / v% if 10 ml is neutralized by 17.5 ml of 0.1 N NaOH ?

c1 * v 1 = c2 * v2

10 ml * x N = 17.5 ml * 0.1 N

x = 0.175 N

Since sulfuric acid is diprotic, its equlivalent mass: 98/2 = 49

0.175 N * 49 = 8.575 g / liter; 0.858 g / 100 ml, 0.858 w/v %

4.1.1.4. What is the degree of dissociation in 0.8 M lactic acid and howdoes it change when it is diluted one hundred times? Ka = 1.4 * 10-4.

1. pH = (pK-lg c) / 2 pK = 3.85; lg c = - 0.097

pH = (3.85 + 0.097) / 2 = 1.97 [H+] = 0.0107; = 0.0107/0.8 = 0.013

2. pH = (3.85 + 2.097) / 2 = 2.973 [H+] = 0.00107; = 0.00107/0.008 = 0.13

= 0.013 in 0.8 N lactic acid, = 0.13 after a hundredfold dilution.

4.1.2.9. 20 ml 10 w / w% sulfuric acid (d = 1.08 g / cm3) is diluted to 5 liter. Calculate the pH !

The w/v % concentration of the solution is 10 * 1.08, therefore 10.8 %. Dilution factor of 20 ml solution to 5 liter is 250 times.If so, the final concentration is 10.8/250 = 0.0432 w/w %.

The molar mass of sulfuric acid is 98, being a diprotic acid, its equivalent mass is 49.

The 0.0432 % concentration corresponds to 0.432 g/liter.

Using these data, the [H+] = 0.432 / 49 = 8.8 * 10-3

pH = - lg [H+] = 0.432 / 49 = 8.8 * 10-3 = 2.05

4.1.1.7. What is the degree of dissociation in a 0.02 N acid solution if Ka = 3 * 10-2?

c α2

Ka = ; therefore c α2 + αKa - Ka = 0 1 – α __________ ___________________ - Ka ± Ka

2 - 4c(-Ka) - 3 * 10-2 + 9*10-4 + 24*10-4

α = = =

2c 4 * 10-2

- 3 * 10-2 + 5.74 * 10 -2

= = 0.685 4 * 10 -2

4.1.2.14. What is the pH in a 0.035 N solution of an organicamine if its pKa = 9.6?

Organic amines are weak bases. pKb = 14 – pKa

Therefore pKb = 4.4

pOH = (pKb – lg c) / 2, for this amine;pOH = (4.4 + 1.46) / 2 = 5.86 / 2 = 2.93

pH = 14 – pOH = 11.07

4.1.2.5. What is the pH of a 0.01 M HCl solution ?

Since HCl is a strong acid, for the first sight the pH of 10-8 N HCl would be 8.However, in this case, the continuous dilution of a strong acid might result in more concentrated strong base.

Where is the error ?When we use the equation [H+] = [strong acid], the protonsderived from the ionization of water are neglected. However, when an acid is very diluted, this is not correct.

In this solution, [H+] from the water is 10-7; [H+] from HCl is 10-8. The total [H+] = 10-7 + 10-8 = 1.1 *10-7

pH = -lg 1.1 *10-7 = 6.96

4.1.2.17. How does the pH of a 0.1 N lactic acid solutionchange if it is diluted hundredfold? Ka = 1.4 * 10-4.

pH = (pKa-lg c) / 2 = (3.85 + 1) / 2 = 2.425

After 100 * dilution, c= 0.001 N

pH = (3.85 + 3) / 2 = 3.425

Therefore pH = + 1

4.1.2.7. What is the pH of a mixture of 50 ml 0.7 M sulfuricacid and 50 ml 1 M NaOH ?

The sulfuric acid is in excess, because it is diprotic (its concentration in normality is 1.4 N).50 ml sulfuric acid contains 0.035 mole H2SO4 = 0.07 M [H+]50 ml 1 M NaOH contains 0.05 M NaOH = 0.05 M [OH-]

After their reaction, 0.02 mole [H+] remains not neutralized.This is dissolved in 100 ml total volume, therefore the [H+]–concentration is 0.2 mole/liter (M). pH = - lg [H+] = - lg 0.2 = 0.7

4.1.2.13. What is the pH and the degree of dissociation in a 1 mM weak acid solution if its Ka = 1.6 * 10-6 ?

pH = (pKa – lg c) / 2

pH = (5.8 + 3) / 2 = 8.8 / 2 = 4.4

[H+] = - invlog pH = 4 * 10-5

α = [H+] / [acid] = 4*10-5 / 10-3 = 0.04 (4 %)

4.1.3.2. Calculate the pH of an acetate buffer containing 0.1 M acetic acid and 0.05 M sodium acetate ! pKa= 4.7

Henderson-Hasselbalch equation general form: [deprotonated] pH = pKa + lg [protonated]

0.05 pH = 4.7 + lg = 4.4 0.1

4.1.3.3. A buffer is composed of 0.25 M ammonia and 0.5 M NH4Cl. 20 ml 0.2 M HCl is added to 100 ml buffer. Calculate the pH change ! pKb = 4.7.

[deprotonated] pH = pKa + lg and pKa = 14- pKb

[protonated] 0.25 [deprot]- [H+] pH = 9.3 + lg = 9.0; pH = pKa + lg 0.5 [prot] + [H+]20 ml 0.2 M HCl contains 0.004 mole HCl. 100 ml buffer contains 0.025 mole ammonia, 0.05 mole NH4Cl

0.025 – 0.004 pH = 9.3 + lg = 8.89 pH = - 0.11 0.05 + 0.004

4.1.3.10. 2 g NaOH is dissolved in 1 liter 0.2 M acetic acid. What is the pH if the pKa= 4.7 ?

2 g NaOH = 0.05 mole; 1 liter 0.2 M acetic acid = 0.2 moleThe acid is in excess; the result is a buffer.

[deprotonated] pH = pKa + lg [protonated]

0.05 M pH = 4.7 + lg = 4.22 0.2 - 0.05 M

4.1.3.14. Calculate the acetic acid and acetate concentrations in a 0.2 M acetate buffer (pH=5.0) ! pKa= 4.7.

0.2 M concentration means that [acetate]+ [acetic acid] = 0.2M

[deprotonated] pH = pKa + lg [protonated]

[acetate] [acetate] 5.0 = 4.7 + lg ; 0.3 = lg 0.2 - [acetate] 0.2 - [acetate]

[acetate] / 0.2 - [acetate] = 2 [acetate] = 2* 0.2 – 2 * [acetate] 3 [acetate] = 2* 0.2 [acetate] = 0.4 /3 = 0.133 M; [acetic acid]=0.2–0.133 = 0.067 M

4.1.3.12. What is the pH of a solution prepared from 57 ml cc.acetic acid (d= 1.05 g / cm3), 14 g NaOH and 33 ml 39 w / w% HCl (d= 1.2 g / cm3) and is filled with distilled water to 1 liter. pKa of acetic acid is 4.7.

The mass of acetic acid is 59.85 g, 0.997 mole.14 g NaOH represents 0.35 mole.33 ml HCl is 15.44 g which represents 0.42 mole

Since HCl + acetic acid are in excess, 0.07 mole [H+] is present (acetic acid is not ionized) – it is not a buffer !

pH = - lg [H+] = - lg 0.07 = 1.15

4.1.3.17. After muscle work load the pH of a blood sample is 7.32. Partial pressure of carbon dioxide is 32 Hgmm, 1 Hgmmis equivalent with 0.03 mM dissolved CO2. Calculate the HCO3

-/CO2 proportion and the total CO2-content expressed in ml/100 ml plasma value at 38oC, if the molarvolume of CO2 is 22.26 liter ?

32 Hgmm dissolved CO2 corresponds to 0.96 mM.

[deprotonated] pH = pKa + lg pKa = 6.1 [protonated] [HCO3

-] [HCO3-] [HCO3

-] 7.32 = 6.1 + lg ; lg = 1.22; = 16.6 [CO2] [CO2] [CO2][HCO3

-] = 16.6 * 0.96 = 15.94 mMTotal CO2 content is 15.94 + 0.96 = 16.9 mM In 100 ml, the total CO2 is 1.69 mmoles, i.e. 37.62 ml / 100 ml

4.1.3.18. During breath holding the total CO2 concentration (CO2+HCO3

- ions) increases to 30 mM and the pressure of CO2 to 60 Hgmm. What is the pH of the blood? 1 Hgmm = 0.03 mM CO2. pKa = 6.1.

60 Hgmm corresponds to 1.8 mM CO2.Consequently, [HCO3

-] = 30-1.8 = 28.2 mM 28.2 pH = 6.1 + lg = 6.1 + 1.19 = 7.29 1.8

4.1.3.21. The carbon dioxide / hydrocarbonate buffer is important in the maintenance of blood pH. Calculate the buffer capacity of 1 liter blood plasma (acid and base is given in mmole amounts !) if the pK of the buffer is 6.1, HCO3- concentration is 28 mM, CO2 concentration is 1.4 mM ? Neglect the effects of other blood buffer systems ! 28 pH = 6.1 + lg = 7.4; before adding acid/base 1.4For acid: 28 – x 28-x 28- x 6.4 = 6.1 + lg ; 0.3 = lg ; = 2 1.4 + x 1.4 + x 1.4 + x 28 – x = 2.8 + 2x; 3 x = 25.2 x = 8.4 mM for acidFor base: 28 + x 28 + x 28 + x 8.4 = 6.1 + lg ; 2.3 = lg ; = 200 1.4- x 1.4 – x 1.4 – x28 + x = 280 – 200 x; 201 x = 252 x = 1.25 mM for base

4.1.4.6. How many ml 1 M NaCl should be added to 10 ml 1 mM AgNO3 to initiate the precipitation ? Solubility productof AgCl is 10-10.

[Ag+] [Cl-] = 10-10; [Cl-] = 10-7

c1 * v1 = c2 v2

10-6 * x = 10-7 * (10 + x) 10-6x = 10-6 + 10-7x 9*10-7x = 10-6, x = 10-6 / 0.9*10-6 = 1.11 ml

4.1.4.4. Calculate the solubility of PbI2 in water if its solubilityproduct is 9 * 10-9 !

Solubility is equal to the concentration of dissolved Pb2+

[Pb2+] [I-]2 = 9 *10-9

In equilibrium 2 [Pb2+] = [I-], i.e. Ksp = [Pb2+] [2Pb2+]2

9 * 10-9 = 4 [Pb2+]3 9 * 10-9 3

[Pb2+]3 = ; [Pb2+] = | 9*10-9 / 4 4

[Pb2+] = 1.31 * 10-3

Thank you for your attention !

Use your „Selected calculation of Chemical Calculations and Biochemical Exercices”

book for the preparation