Acids and Bases Acids and Bases
The Danish chemist Johannes Brnsted and the English chemist A common equilibrium situation involves acids and bases. Thomas Lowry put forward a model:
ACID: proton donor (H+) Many cell processes rely on the correct pH balance. BASE: proton acceptor The charge state, and hence solubility, of many biological molecules is pH dependant. In this context we are always looking at aqueous solutions in which the ‘proton’ is extensively hydrated. Acids were originally recognized by their sour taste. + + It is represented as H (aq) or H3O (aq). Bases often have a bitter taste H and slippery feel H HOH HOH
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Conjugate Pair Question
+ A conjugate acid - base pair differ by H Write the formula of the Write the formula of the + e.g. NH4 is the conjugate acid of NH3 conjugate bases conjugate acids + NH3 is the conjugate base of NH4 + – H3O H2O OH H2O H H O H Cl H O Cl – + H SO HSO H O H3O H H 2 4 4 2 Base Acid Acid Base – – HClO4 ClO4 Cl HCl
– + CH COOH CH COO NH NH4 H H 3 3 3
HO N HO H N H H H H H
AcidBase Base Acid 3 4
1 Strong & weak acids & bases pH Scale
+ Strong acids and bases completely dissociate in water and [H ] is used as a measure of acidity and the pH of the solution is calculated directly from the for convenience a log scale is employed concentration of the acid. pH = - log [H+ (aq)] Common strong acids include: pOH = - log [OH- (aq)]
HCl, HBr, HI, HNO3, H2SO4 Common strong bases include: In acid solution (pH < 7) [H+] > 1x10-7 M NaOH, KOH, Ca(OH) 2 In basic solution (pH > 7) [H+] < 1x10-7 M
pH and pOH are interconverted easily Weak acids and bases are in an equilibrium and so it is common to refer to just pH - + - even when we have high [OH ]. HA (aq) + H2O H3O (aq) + A (aq) + - [H (aq)][A (aq)] pH + pOH = 14 + - Ka = in short: HA (aq) H (aq) + A (aq) [HA(aq)] 5 6
Examples Neutral solution
In strong acids and bases, complete dissociation occurs + - and pH calculated directly from starting concentrations H2O H (aq) + OH (aq)
+ - -14 2 Calculate the pH of: Kw = [H ][OH ] = 1x10 M at 25 °C
0.001 M HNO3 [H+] = 0 .001 M , pH30H = 3.0 In neutral solution [H+] = [OH-] = 1x10-7 M
0.001 M NaOH and pH = 7.0 (25 C) [OH -] = 0.001 M, pOH = 3.0, pH = 11.0
0.001 M Ca(OH)2 [OH -] = 0.002 M, pOH = 2.7, pH =11.3 7 8
2 Question Calculations using Ka
-14 2 -5 K is dependant on temperature. At 37 C Kw = 2.45 x 10 M . Calculate the pH of acetic acid CH3COOH (0.50 M). Ka = 1.76 x 10 M Calculate the pH of a neutral solution at physiological + - temperature. CH3COOH H + CH3COO initial conc / M 0.50 - - 1. 6.4 change / M -x x x
2. 6.8 equilibrium conc / M 0.50 - x x x
3. 7.0 + - -5 2 Ka = [H ][CH3COO ] = 1.76 x 10 M = x / (0.50 - x) 4. 7.2 [CH3COOH]
5. 7.4 assume x << 0.5 then x2 = (1.76 x 10-5)(0.50) = 8.76 x 10-6 0% 0% 0% 0% 0% x = [H+(aq)] = 2.96 x 10-3 M 6.4 6.8 7 7.2 7.4 + -3 9 pH = -log[H ] = -log[2.96 x 10 ] = 2.53 10
Question Polyprotic acids
Have more than one proton available. Benzoic acid is used as a preservative in solutions and in mouth wash. A bottle of Listerine indicates the concentration of benzoic acid used is 1.5 mg mL-1, what is Example: H CO (carbonic acid) has 2 protons that can be -5 2 3 the pH? (C6H5COOH, Molar mass = 122, Ka =6.3 x 10 M) donated stepwise, one at a time - diprotic acid. + - -7 H2CO3 H + HCO3 Ka1 = 4.3 x 10 M 1.) 1.5 mg mL-1 is equivalent to 1.5 g L-1, [C H COOH] = 1.5/122 = 0.012 M - + 2- -11 6 5 HCO3 H + CO3 Ka2 = 5.6 x 10 M
+ - 2.) C6H5COOH H + C6H5COO Phosphoric acid is a triprotic acid. + - -3 initial conc / M 0.012 - - H3PO4 H + H3PO4 Ka1 = 7.5 x 10 M - + 2- -8 change / M -x x x H2PO4 H + HPO4 Ka2 = 6.2 x 10 M equilibrium conc / M 0.012 - x x x 2- + 3- -13 HPO4 H + PO4 Ka3 = 4.8 x 10 M
+ - -5 2 Ka = [H ][C6H5COO ] = 6.3 x 10 M = x / (~0.012) For a typical weak polyprotic acid K > K > K indicating [C6H5COOH] a1 a2 a3 x2 = (6.3 x 10-5)(0.012) that the loss of the first proton is always much easier than x = [H+(aq)] = 0.00087 M subsequent ones. pH = -log[H+] = -log[0.00087] = 3.1 11 12
3 pKa and pKb Buffer solutions pH = - log[H+] A buffer solution withstands changes in pH when a limited amount of acid or base is added. pOH = - log[OH-] and pH + pOH = 14 It is composed of a weak acid and a salt of a weak acid
with a strong base (e.g. CH3COOH + CH3COONa) or a Similarly pKa = - log Ka weak base and a salt of a weak base with a strong acid and pKb = - log Kb and pKa + pKb = 14 (e.g. NH3 + NH4Cl).
The smaller pKa, the stronger the acid.
The smaller pKb, the stronger the base.
Usually pKa is reported for acids and bases – for bases it refers to the conjugate pair.
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Charge and pH Charge and pH
This is how to work out the charge on a molecule at a given pH Many biological molecules are sensitive to pH . When pH = pKa, then [acid] = [base]. . Proton transfer is involved in many reactions. . When pH < pK , then protonated form dominates. . The charge on a molecule containing acid & base groups a depends on the pH. . When pH > pKa, then de-protonated form dominates. . Salivary amylase starts the H H O digestion of starches in the H N C C mouth and has an optimum H CH2 O activity at pH = 6.8 H at pH = 3.0 Example: Glutamic acid pKa=9.67 O CH pKa=2.19 2 . Pepsin starts the digestion of has the following H N C C proteins in the stomach C H CH2 OH O O H which is very acidic. It has structure; what is the CH2 an optimum activity at pH = charge on the molecule 2.0 C pKa=4.25 H H O O OH . Protein digestion continues at pH 3.0 and pH 8.5? H N C C in the small intestine where at pH = 8.5 H CH2 O the pancreas releases CH2 trypsin which has an C optimum activity at pH = 9.5 O O 15 16
4 Question Question
What is the overall H O pKa=2.18 If you wished to administer charge on the basic pKa=8.95 H3N C C CH2 O a water soluble form of amino acid lysine at CH2 mophine (pKa (amine) ~ pH 7.4? CH2 CH2 9.4) to an animal, what pH pKa=10.53 NH3 would you use? 1. 2-
2. 1- 1. pH = 11.4 3. no charge 2. pH = 9.4 4. 1+ 5. 2+ 0% 0% 0% 0% 0% 3. pH = 7.4 2- 1- 1+ 2+ 4. pH = 4.4 0% 0% 0% 0%
pH = pH = pH = pH = 11.4 9.4 7.4 4.4 no charge no 17 18
Solutions Factors Affecting Solubility
In order for one substance to dissolve in another: In a solution a solute dissolves homogenously in a solvent Solute molecules must separate from each other Usually there is much more solvent than solute Solvent molecules must make room for the The solubility of a substance is the maximum that can ions/molecules of the solute
dissolve in a fixed amount of a particular solvent at a Solute and solvent molecules must mix certain temperature
Whether a substance dissolves Solubility of NaCl in 100 mL water at 100 C is 39.12 g or not will depend on relative Solubility of AgCl in 100 mL water at 100 C is 0.0021 g strengths of solute-solute, solute-solvent and solvent-solvent intermolecular forces 19 20
5 Structure and Pressure Effects Henry’s Law p = kC
Structure Effects: Solubility is favoured if the solute and the The relationship between the gas pressure and the solvent have similar polarities. concentration of dissolved gas is given by Henry’s Law
Thus the structure of the . p is the partial pressure of the gas above the solution molecules plays a definite part . C represents the concentration of the dissolved gas in the solubility of compounds. . k itthtitiftilis a constant characteristic of a particular gas-soltilution
The smaller the value of k the more SOLUBLE the gas Pressure Effects: Gases - Carbonated beverages are always EXAMPLE: Gas k Helium 2700 atm M-1 bottled at high pressures of CO2 to -1 ensure a high concentration of CO2 in Nitrogen 1600 atm M the drink Carbon dioxide 29 atm M-1
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Deep Sea Diving Osmotic Pressure
Helium used instead of nitrogen mixed The flow of solvent molecules into a solution through a semi-permeable membrane is called OSMOSIS. with O2 in deep sea diving At significant depths (eg 30 m) the air in the It is the number of particles in lungs is around 4 atm, which causes a the sooulutio n which is ltfNlarge amount of N andOd O tdilto dissolve i ithbldOn the blood. O can be 2 2 2 important, not their chemical metabolised but as the diver ascends, if done too quickly, the N2 cannot be released quickly enough and so the diver suffers ‘the nature. bends’.
He is ~half as soluble as nitrogen and so less must be released More solvent molecules enter avoiding this problem. the solution than leave it. As a He also has a higher rate of diffusion in the blood which makes He more acceptable. result the solution volume increases and its concentration decreases. 23 24
6 Osmotic Pressure Osmotic Pressure
The solution exerts a = cRT backward pressure - the osmotic pressure () that eventually equalises the flow of solvent molecules in is the osmotic pressure in atm, both directions -1 c is the molarity of the solute (mol L )
The osmotic pressure is R is the gas law constant and also defined as the applied pressure required to T is the temperature in K prevent this volume change
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Example Isotonic Solutions
Contact lens rinses consist of 0.015 M NaCl solution to ISOTONIC SOLUTION -identical osmotic prevent any changes in volume of corneal cells. pressure to cell fluid What is the osmotic pressure of this saline solution at 38 C? HYPOTONIC SOLUTIONS –lower osmotic pressure than cell fluid - cell swells & ruptures - = cRT transfer of fluid INTO CELL – LYSIS
= (2 x 0.015 mol L-1)(0.082 L atm K-1 mol-1)(273 + 38 K) HYPERTONIC SOLUTIONS -higher osmotic pressure than cell fluid - cell shrivels - transfer of = 0.77 atm fluid OUT OF CELL – CRENATION
The word “tonocity” refers to the tone, or firmness of a biological cell. 27 28
7 Question
Solutions for the intravenous injection must have the same osmotic pressure as blood. According to the label on the bottle each 100 mL of normal saline solution used for intravenous injection contains 900 mg of NaCl.
What is the molarity of NaCl in normal saline solution? Amount (NaCl) = 0.900 g/58.44 g mol -1 = 0.0154 mol [NaCl] = 0.0154 mol / 0.100 L = 0.154 mol L-1
What is the osmotic pressure of normal saline solution at the average body temperature of 37 C? = cRT = (2 x 0.154 mol L-1)(0.082 L atm K-1 mol-1)(310 K) = 7.83 atm ~ 8 atm (to 1 sig fig)
What concentration of glucose (C6H12O6) has the same osmotic pressure as normal saline solution. [glucose] = 2 x 0.154 mol L-1 = 0.308 mol L-1 29
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