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Basics of Medical Chemistry Course 5. lecture. Consultation for solution of calculation

Basics of Medical Chemistry Course 5. lecture. Consultation for solution of calculation problems – preparation for the 1st Midterm Concentrations Titration problems Ionization, pK, degree of dissociation pH of strong and weak acids Buffers Solubility product, solubility Andras Hrabak

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Basics of Medical Chemistry Course 5. lecture. Consultation for solution of calculation

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  1. Basics of Medical Chemistry Course • 5. lecture. Consultation for solution of calculation • problems – preparation for the 1st Midterm • Concentrations • Titration problems • Ionization, pK, degree of dissociation • pH of strong and weak acids • Buffers • Solubility product, solubility • Andras Hrabak • Department of Medical Chemistry, Molecular Biology and • Pathobiochemistry • 6 October 2009

  2. 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

  3. 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. 4.1.1.4. What is the degree of dissociation in 0.8 M lactic acid and how does 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.

  5. 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

  6. 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± Ka2 - 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

  7. 4.1.2.14. What is the pH in a 0.035 N solution of an organic amine 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

  8. 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 protons derived 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

  9. 4.1.2.17. How does the pH of a 0.1 N lactic acid solution change 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

  10. 4.1.2.7. What is the pH of a mixture of 50 ml 0.7 M sulfuric acid 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

  11. 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 %)

  12. 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

  13. 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

  14. 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 mole The 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

  15. 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

  16. 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

  17. 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 Hgmm is 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 molar volume 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 mM Total 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

  18. 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

  19. 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 baseis 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 1.4 For 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 acid For base: 28 + x 28 + x 28 + x 8.4 = 6.1 + lg ; 2.3 = lg  ;  = 200 1.4- x 1.4 – x 1.4 – x 28 + x = 280 – 200 x; 201 x = 252 x = 1.25 mM for base

  20. 4.1.4.6. How many ml 1 M NaCl should be added to 10 ml 1 mM AgNO3 toinitiate the precipitation ? Solubility product of 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

  21. 4.1.4.4. Calculate the solubility of PbI2 in water if its solubility product 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 * 10-93 [Pb2+]3 =  ; [Pb2+] = | 9*10-9 / 4 4 [Pb2+] = 1.31 * 10-3

  22. Thank you for your attention ! Use your „Selected calculation of Chemical Calculations and Biochemical Exercices” book for the preparation

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