pH and Buffers Calculations PDF

pH and Buffers  pH and B uffers Protons (H+) are commonly released when some cellular components are dissolved in water. Cellular components can either be organic or inorganic, such as weak acids. These acids, when in aqueous solutions release H+ ions and coexist...

pH and Buffers  pH and B uffers Protons (H+) are commonly released when some cellular components are dissolved in water. Cellular components can either be organic or inorganic, such as weak acids. These acids, when in aqueous solutions release H+ ions and coexists in equilibrium with their conjugate bases. The coexistence of these acids and their corresponding conjugates, together with other dissolved salts maintain regulated changes in the H+ ion concentration and constitute the buffer systems of cells.  pH and B uffers A cell produces H+ ions in amount equivalent to 2 L of HCl over a 24-hour period during breaking down of substances. Increased H+ concentration can drastically lower the cell’s pH to less than the physiological conditions. Cell’s physiological pH is from 6.8 to 7.4. A very weak acid, such as water dissociates into However, its Kw is low and its [H+] and [OH-] is only 10-7 and can’t be able to neutralize even small amounts of added acids or bases. To ensure the maintenance of the pH of cells within the physiological range, presence of weak acids in equilibrium with its conjugate bases or dissolved salts are responsible for this action.  pH and B uffers To ensure the maintenance of the pH of cells within the physiological range, presence of weak acids in equilibrium with its conjugate bases or dissolved salts are responsible for this action. Dextrose and sodium chloride solutions are used as sources of electrolytes, calories, and water for hydration. Sodium and chloride ions are responsible for regulating the acid-base balance of the body. While dextrose is a source of calories. K W , ION P RODUCT CONSTANT OF WATER  The pH Scale pH is the negative logarithm of the hydrogen ion concentration.  The pH Scale The pH scale is widely used in biological applications. Hydrogen ion concentrations in biological fluids are very low, about 10-7 M (0.0000001 M), a value more easily represented as pH=7. For example the blood plasma, pH=7.4 (0.00000004 M) The pOH Scale  We ak E l ectrolytes A r e Su bstances Th at D i ssociate O n ly Sl ightly I n Wa ter Substances with only a slight tendency to dissociate to form ions in solution are called weak electrolytes. Acetic acid, CH3COOH, is a good example: The acid dissociation constant, Ka for acetic acid is 1.74 x 10-5 M: Ka, is also termed an ionization constant because it states the extent to which a substance forms ions in water.  Buffers The human body’s physiological pH is about 6.8 to 7.4, the pH that most chemical reactions occur in the body. It is also at this pH that the cells, tissues and organs carry out their functions in the body. Chemical reactions are carried out by the cell like breaking down of substances produces H+ ions, which may increase as the reaction progresses. This may lead to the lowering of the blood pH known as acidosis. There may also be reactions which may lead to the increase in blood pH that leads to alkalosis. Such conditions could be detrimental to an organism. Thus buffers are present in biochemical systems to prevent damage that would result from an increase in acid or alkali production.  Buffers (Mathews 4th Ed.)  EXAMPLE 1. Determine the pH of a solution if [OH -] = 0.040 M. 𝐩𝐎𝐇 = −𝐥𝐨𝐠 𝑶𝑯− pOH = −log 0.040 = 1.3979 𝐩𝐇 + 𝐩𝐎𝐇 = 𝟏𝟒 pH = 14 − pOH pH = 14 − 1.3979 𝐩𝐇 = 𝟏𝟐. 𝟔𝟎 (final answer) EXAMPLE 1 1. Press – (minus) sign Just input the values 2. Press log and press = 3. Input the value 0.040 4. Press ) 4. Press = EXAMPLE 1 (SHORTCUT) 1. Input 14 2. Press – sign 3. Press ( 4. Press – sign again 5. Press log 6. Input 0.040 value 7. Press ) twice 7. Press =  EXAMPLE 2. Compute the [H+] of a solution when pOH=10.0 𝐩𝐇 + 𝐩𝐎𝐇 = 𝟏𝟒 pH = 14 − pOH pH = 14 − 10.0 pH = 4 𝐩𝐇 = −𝐥𝐨𝐠 𝐇 + −pH = log H + antilog(−pH) = H + 𝐇 + = 𝟏𝟎−𝐩𝐇 H + = 10−4 𝑯+ = 𝟏 𝐱 𝟏𝟎−𝟒 𝐨𝐫 𝟎. 𝟎𝟎𝟎𝟏 𝐌 (final answer) EXAMPLE 2 Just input the values 1. Press Shift and press = 2. Press log 3. Press – 4. Input the 4 value 5. Press = EXAMPLE 2 (SHORTCUT) 1. Press Shift 2. Press log 3. Press – 4. Press ( 5. Input 14 6. Press – 7. Input 10 8. Press ) 9. Press =  EXAMPLE 3. Calculate the pH of: 5.0 x 10-4 M HCl 1. Press – 2. Press log 𝐩𝐇 = −𝐥𝐨𝐠 𝑯+ 3. Input 5 x 10 pH = −log 5.0 x 10−4 4. Press 5. Input –4 𝐩𝐇 = 𝟑. 𝟑𝟎 (final answer) 6. Press 7. Press ) 8. Press =  EXAMPLE 4. Calculate the pH of: 7.0 x 10-5 M NaOH 𝐩𝐎𝐇 = −𝐥𝐨𝐠 𝑶𝑯− Practice using the pOH = −log 7.0 x 10 −5 shortcut method: 1. Press 14 pOH = 4.1549 2. Press – 3. Press ( 4. Press – again 𝐩𝐇 + 𝐩𝐎𝐇 = 𝟏𝟒 5. Press log pH = 14 − pOH 6. Input 7 x 10 7. Press pH = 14 − 4.1549 8. Input – 5 𝐩𝐇 = 𝟗. 𝟖𝟓 (final answer) 9. Press 10. Press ) twice 11. Press =  EXAMPLE 5. Calculate the pH of: 3.0 x 10-2 M KOH 𝐩𝐎𝐇 = −𝐥𝐨𝐠 𝑶𝑯− pOH = −log 3.0 x 10 −2 Practice using the pOH = 1.5229 shortcut method from the previous slide. 𝐩𝐇 + 𝐩𝐎𝐇 = 𝟏𝟒 pH = 14 − pOH pH = 14 − 1.5229 𝐩𝐇 = 𝟏𝟐. 𝟒𝟖 (final answer)  EXAMPLE 6. Calculate the [H+] of saliva given in Table 2.3: saliva, pH = 6.6 𝐩𝐇 = −𝐥𝐨𝐠 𝑯+ −pH = log H + antilog(−pH) = H + 𝐇 + = 𝟏𝟎−𝐩𝐇 H + = 10−6.6 𝑯+ = 𝟐. 𝟓𝟏 𝐱 𝟏𝟎−𝟕 𝐌 (final answer) EXAMPLE 6 1. Press Shift 2. Press log 3. Press – 4. Input the 6.6 value 5. Press =  EXAMPLE 7. Calculate the [OH-] in milk of magnesia given in Table 2.3: Milk of magnesia, pH = 10.3 𝐩𝐇 + 𝐩𝐎𝐇 = 𝟏𝟒 pOH = 14 − pH pOH = 14 − 10.3 pOH = 3.7 𝐩𝐎𝐇 = −𝐥𝐨𝐠 𝐎𝑯− −pOH = log OH − antilog(−pOH) = OH − 𝐎𝐇 − = 𝟏𝟎−𝐩𝐎𝐇 OH − = 10−3.7 𝐎𝐇 − = 𝟏. 𝟗𝟗𝟓 𝐱 𝟏𝟎−𝟒 𝐌 (final answer) EXAMPLE 7 1. Press Shift 2. Press log 3. Press – 4. Press ( 5. Input 14 6. Press – 7. Input 10.3 8. Press ) 9. Press =  EXAMPLE 8. Calculate the [H+] inside a liver cell given in Table 2.3: Liver, pH = 6.9 𝐩𝐇 = −𝐥𝐨𝐠 𝑯+ −pH = log H + antilog(−pH) = H + 𝐇 + = 𝟏𝟎−𝐩𝐇 H + = 10−6.9 𝐇 + = 𝟏. 𝟐𝟔 𝐱 𝟏𝟎−𝟕 𝐌 (final answer)  EXAMPLE 9. Hydrochloric acid is a significant component of gastric juice. What is the concentration of hydronium ion (H3O+ or H+) in gastric juice if pH=1.2? Gastric juice, pH = 1.2 𝐩𝐇 = −𝐥𝐨𝐠 𝑯+ −pH = log H + antilog(−pH) = H + 𝐇 + = 𝟏𝟎−𝐩𝐇 H + = 10−1.2 𝐇 + = 𝟎. 𝟎𝟔𝟑𝟏 𝐌 (final answer)  Buffers A solution that resists drastic changes in pH when small amounts of acid or base is added to it. A buffer system consists of a weak acid (proton donor) and its conjugate base (proton acceptor). Buffers are amphoteric substances, that is, they contain both acidic and basic groups. Buffer systems in cell: Weak acids and its conjugate base Amino acids  Buffers Buffers formed from a weak acid and its conjugate base. Consider the dissociation of a weak acid, acetic acid: pKa = 4.7  Buffering Capacity Buffer capacity: The amount of hydronium or hydroxide ions that a buffer can absorb without a significant change in pH. Buffer capacity depends both its pH and its concentration. Maximum buffering capacity pH = +1/-1 unit of the pKa of the weak acid. Example if acetic acid its pKa = 4.75, therefore acetic acid/sodium acetate function as an effective buffer within a pH range of approximately 3.75-5.75. Note: pKa is the acid dissociation constant of a solution, it is a method of determining an acid's strength, hence, a lower pKa value denotes a strong acid (Mathews 4th Ed.)  Henderson-Hasselbalch Equation Henderson-Hasselbalch equation: A mathematical relationship between: pH, pKa of the weak acid HA and concentrations of HA and its conjugate base A-. A convenient way to calculate the pH of a buffer when the concentrations of the weak acid and its conjugate base are not equal. It is derived in the following way: taking the logarithm of this equation gives:  Henderson-Hasselbalch Equation Multiplying by -1 gives: -log Ka is by definition pKa, and -log [H3O+] is by definition pH. Making these substitutions gives: rearranging terms gives:  Solving Problems with Henderson-Hasselbalch Equation 1. Calculate the pKa of lactic acid, given that when the concentration of free lactic acid is 0.010 M and concentration of lactate is 0.087 M, the pH is 4.80. [𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒] pH = pKa + log [𝑤𝑒𝑎𝑘 𝑎𝑐𝑖𝑑]  Solving Problems with Henderson-Hasselbalch Equation 1. Calculate the pKa of lactic acid, given that when the concentration of free lactic acid is 0.010 M and concentration of lactate is 0.087 M, the pH is 4.80. [𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒] pH = pKa + log [𝑤𝑒𝑎𝑘 𝑎𝑐𝑖𝑑] [𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒] pKa = pH – log [𝑤𝑒𝑎𝑘 𝑎𝑐𝑖𝑑] 0.087 M pKa = 4.80 - log 0.010 M pKa = 4.80 – log 8.7 pKa = 3.86 (final answer) S olving Pr oblems with H ender s on -H as selbalch Equation 1 1. Input 4.8 2. Press – 3. Press log 4. Press 5. Input the value 0.087 in the numerator 6. Press 7. Input the value 0.01 in the denominator 8. Press 9. Press ) 10. Press =  Solving Problems with Henderson-Hasselbalch Equation 2. Calculate the pH of a mixture of 0.1 M acetic acid and 0.2 M sodium acetate. The pKa of acetic acid is 4.76. [𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒] pH = pKa + log [𝑤𝑒𝑎𝑘 𝑎𝑐𝑖𝑑]  Solving Problems with Henderson-Hasselbalch Equation 2. Calculate the pH of a mixture of 0.1 M acetic acid (CH3COOH) and 0.2 M sodium acetate (CH3COONa). The pKa of acetic acid is 4.76. [𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒 ] pH = pKa + log [𝑤𝑒𝑎𝑘 𝑎𝑐𝑖𝑑] [𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒] pH = pKa + log [𝑤𝑒𝑎𝑘 𝑎𝑐𝑖𝑑] Practice on your own 0.2 M using the step by pH = 4.76 + log 0.1 M step process in pH = 4.76 + 0.301 Problem No.1 pH = 5.06 (final answer)  Solving Problems with Henderson-Hasselbalch Equation 3. What is the pH of a buffer if 0.1 M acetic acid (CH3COOH) and 0.2 M sodium acetate (CH3COONa) are present. The Ka of acetic acid is 1.8 x 10-5. pKa = −log Ka pKa = −log 1.8 × 10−5 pKa = 4.7447 conjugate base pH = pKa + log weak acid 0.2 pH = 4.7447 + log 0.1 pH = 4.7447 + log 2 pH = 4.7447 + 0.3010 𝐩𝐇 = 𝟓. 𝟎𝟓 (final answer) S olving Pr oblems with H ender s on -H as selbalch Equation 3 Originally; 𝐜𝐨𝐧𝐣𝐮𝐠𝐚𝐭𝐞 𝐛𝐚𝐬𝐞 𝐩𝐇 = 𝐩𝐊𝐚 + 𝐥𝐨𝐠 𝐰𝐞𝐚𝐤 𝐚𝐜𝐢𝐝 Since; 𝐩𝐊𝐚 = −𝐥𝐨𝐠 𝐊𝐚 Then; 𝐜𝐨𝐧𝐣𝐮𝐠𝐚𝐭𝐞 𝐛𝐚𝐬𝐞 𝐩𝐇 = −𝐥𝐨𝐠 𝐊𝐚 + 𝐥𝐨𝐠 𝐰𝐞𝐚𝐤 𝐚𝐜𝐢𝐝 Practice on your own and try to input this shortcut method in you calculator. −𝟓 𝟎. 𝟐 − 𝐥𝐨𝐠 𝟏. 𝟖 × 𝟏𝟎 + 𝐥𝐨𝐠 𝟎. 𝟏 Then press = to get the correct answer.  Solving Problems with Henderson-Hasselbalch Equation 4. Calculate the ratio of the concentrations of acetate and acetic acid required in a buffer system of pH 5.30. [𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒] pH = pKa + log [𝑤𝑒𝑎𝑘 𝑎𝑐𝑖𝑑]  Solving Problems with Henderson-Hasselbalch Equation 4. Calculate the ratio of the concentrations of acetate and acetic acid required in a buffer system of pH 5.30. [𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒 ] pH = pKa + log [𝑤𝑒𝑎𝑘 𝑎𝑐𝑖𝑑] [𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒] log = pH - pKa [𝑤𝑒𝑎𝑘 𝑎𝑐𝑖𝑑] [𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒] log [𝑤𝑒𝑎𝑘 𝑎𝑐𝑖𝑑] = 5.30 – 4.76 [𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒] log [𝑤𝑒𝑎𝑘 𝑎𝑐𝑖𝑑] = 0.54 [𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒 ] = antilog 0.54 [𝑤𝑒𝑎𝑘 𝑎𝑐𝑖𝑑] [𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒 ] [𝑤𝑒𝑎𝑘 𝑎𝑐𝑖𝑑] = 100.54 [𝒄𝒐𝒏𝒋𝒖𝒈𝒂𝒕𝒆 𝒃𝒂𝒔𝒆] = 3.47 (final answer) [𝒘𝒆𝒂𝒌 𝒂𝒄𝒊𝒅]  Solving Problems with Henderson-Hasselbalch Equation 5. Determine the conjugate base and weak acid concentration ratio in a buffer consisting of carbonic acid and ammonium carbonate at pH 6.8. Ka = 4.5 x 10-7 pKa = −log Ka pKa = −log 4.5 × 10−7 pKa = 6.3468 conjugate base conjugate base pH = pKa + log log = 0.4532 weak acid weak acid conjugate base conjugate base pH − pKa = log = antilog 0.4532 weak acid weak acid conjugate base conjugate base 6.8 − 6.3468 = log = 100.4532 weak acid weak acid conjugate base 𝐜𝐨𝐧𝐣𝐮𝐠𝐚𝐭𝐞 𝐛𝐚𝐬𝐞 0.4532 = log = 𝟐. 𝟖𝟒 (final answer) weak acid 𝐰𝐞𝐚𝐤 𝐚𝐜𝐢𝐝 S olving Pr oblems with H ender s on -H as selbalch Equation 3 Originally; 𝐜𝐨𝐧𝐣𝐮𝐠𝐚𝐭𝐞 𝐛𝐚𝐬𝐞 𝐩𝐇 = 𝐩𝐊𝐚 + 𝐥𝐨𝐠 𝐰𝐞𝐚𝐤 𝐚𝐜𝐢𝐝 𝐜𝐨𝐧𝐣𝐮𝐠𝐚𝐭𝐞 𝐛𝐚𝐬𝐞 𝐥𝐨𝐠 = 𝐩𝐇 − 𝐩𝐊𝐚 𝐰𝐞𝐚𝐤 𝐚𝐜𝐢𝐝 𝐜𝐨𝐧𝐣𝐮𝐠𝐚𝐭𝐞 𝐛𝐚𝐬𝐞 = 𝐚𝐧𝐭𝐢𝐥𝐨𝐠 (𝐩𝐇 − 𝐩𝐊𝐚) 𝐰𝐞𝐚𝐤 𝐚𝐜𝐢𝐝 𝐜𝐨𝐧𝐣𝐮𝐠𝐚𝐭𝐞 𝐛𝐚𝐬𝐞 = 𝟏𝟎𝐩𝐇−𝐩𝐊𝐚 𝐰𝐞𝐚𝐤 𝐚𝐜𝐢𝐝 Since; 𝐩𝐊𝐚 = −𝐥𝐨𝐠 𝐊𝐚 Then; 𝐜𝐨𝐧𝐣𝐮𝐠𝐚𝐭𝐞 𝐛𝐚𝐬𝐞 = 𝟏𝟎𝐩𝐇−(−𝐥𝐨𝐠 𝐊𝐚 ) 𝐰𝐞𝐚𝐤 𝐚𝐜𝐢𝐝 S olving Pr oblems with H ender s on -H as selbalch Equation 3 Practice on your own and try to input this shortcut method in you calculator. 𝟔.𝟖 − −𝐥𝐨𝐠 𝟒.𝟓 ×𝟏𝟎−𝟕 𝟏𝟎 Then press = to get the correct answer. Measuring pH (Silberberg 5th Ed.)  B uffer Systems The pH range of human blood is close to 7.35-7.45. Any change greater than 0.10 pH unit in either direction can cause illness. To maintain this pH, the body uses three buffer systems: Carbonate buffer: Weak acid: Carbonic acid (H2CO3) and its; Conjugate base: Bicarbonate (HCO3-) Phosphate buffer: Weak acid: Phosphoric acid (H3PO4) and its; Conjugate base: Hydrogen phosphate (HPO42-) Protein buffers  Buffer Systems Two buffer systems act to maintain intracellular pH essentially constant: phosphate (HPO42-/ H3PO4) system and the histidine system. The pH of the extracellular fluid that bathes the cells and tissues of animals is maintained by the bicarbonate/carbonic acid (HCO3- /H2CO3) system.  H i stidine –Imidazole B u f fer S y stem Dissociation of the histidine–imidazole group also serves as an intracellular buffering system. Histidine is one of the 20 naturally occurring amino acids commonly found in proteins and possesses as part of its structure an imidazole group, a five-membered heterocyclic ring possessing two nitrogen atoms. Protein-bound and dipeptide histidine may be the dominant buffering system in some cells. Imidazole pKa value increases substantially when combined with other amino acids, as in proteins or dipeptides.