Acid – Base Theories
Various theories have been proposed to define acids and bases some of which include: a) Arrhenius Theory: Arrhenius defined an acid as a substance which in aqueous solution produces hydrogen ions, H+, while a base is a substance which in aqueous solution produces hydroxides ions, OH-. These definitions are unsatisfactory because they only apply to aqueous solutions. E.g. HCl(aq) H+(aq) + Cl-(aq) NaOH(aq) Na+(aq)+ OH-(aq) b) Bronsted-Lowry Theory: In order to widen the scope of acids and bases to include non-aqueous systems, Bronsted and Lowry independently suggested that an acid is a proton donor while a base is a proton acceptor. Acid Base + H+ An acid and a base that are related by the exchange of a proton form an acid-base conjugate pair. A conjugate base is formed when an acid loses a proton in solution while a conjugate acid is a species formed when a base accepts a proton. Acid 1 + Base 2 Base 1 + acid 2 HCl + H2O Cl- + H3O+ HSO4- + H2O SO4- + H3O+ H2O + NH3 OH- + NH3- Acid 1/base 1, base 2/ acid 2, conjugate pairs: HCl/Cl-, H3O+/H2O, HSO4-/SO42- c) Lewis Theory: According to Lewis, an acid is a substance that can accept a lone pair of electrons to form a covalent bond while a base is a substance that donates a lone pair of electrons to form a covalent bond, e.g. Examples of Lewis bases include: NH3, KOH, OH-, CN-, Cl-, H2O, etc (Nucleophiles), Examples of Lewis acids include: H+, Al3+, Fe3+, AlCl3, FeCl3, BF3, etc (Electrophiles).
THE IONIC PRODUCT OF WATER (Kc)
Pure water behaves as an acid and as a base, i.e. it is amphoteric or amphiprotic H2O(l) H3O+(aq) + OH-(aq) Applying the equilibrium law, gives + – Kc = [H3O ][OH ] [H2O]2 Kc[H2O]2 = [H3O+]+ [OH-]. But since the concentration of water in its pure state is constant, Kc[H2O]2 = constant = Kw Therefore, Kw = [H3O+][OH-] or simply Kw = [H+][OH-] Kw is called the ionic product of water. Definition: The ionic product of water is the product of the concentration of hydrogen ions and hydroxide ions from the self-ionization of water at a given temperature. At 250C, [H+] = [OH-] = 1×10-7moldm-3 Therefore, Kw = (1×10-7moldm-3)2 = 1×10-14mol2dm-6 Kw increases with temperature.
pH SCALE
This is a scale that defines pH as the negative logarithm to the base 10 of the hydrogen ion concentration, i.e. 𝑝𝐻 = -log10[H+] = log [H+] From this, pH increases as the concentration of H+ decreases. In pure water at 250C, [H+] = [OH-] and pH = -log10(1×10-7moldm-3) = 7 For an acidic solution [H+] > [OH-] [H+] = 1×10-7 pH < 7 For an alkaline solution [H+] < [OH-] [H+] < 1×10-7 pH > 7 For a neutral solution [H+] = [OH-] = 1×10-7 Therefore, Kw = [OH-][H+] = 1×10-14 -log[OH-] + -log[OH+] = 14 pH + pOH = 14 pH = 14 – pOH and pOH = 14-pH
CALCULATION OF pH FOR STRONG ACIDS AND ALKALINE SOLUTIONS
Examples 1) Calculate the pH of the following solutions: i) 0.5M HCl, ii) 10-14 moldm-3 H2SO4, iii) 0.025M NaOH iv) 25cm3 of 0.16M NaOH is added to 50cm3 of 0.1M HCl. What is the pH of the mixture? v) What is the [H+] for a solution whose 𝑝𝐻 is a) 3.4, b) 7.7
Solutions
i) HCl(aq) H+(aq) + Cl-(aq), pH = -log[H+(aq)] 0.1M 0.1M pH = -log10-1 = 1 ii) [H2SO4] 2 H+ + SO42-, pH = -log[H+(aq)] [H+] = 2(10-4)M, pH = -log[2(10)-4] =3.69 iii) NaOH(aq) Na+(aq) + OH-(aq), pOH = pH = -log[H-(aq)] 0.025M 0.025M pH = -log0.025 = 2 But pH + pOH = 14 pH = 14 – pOH 14 – 2 = 12.0 iv) NaOH(aq) + HCl(aq) NaCl(aq)+ H2O(aq) 1 1 Number of moles of NaOH = molarity (vol.dm-3) = 0.16x25x10-3 = 4.0×10-3 = 0.004moles Number of moles of HCl(aq) = 0.1x50x10-3 = 0.005moles Excess of moles of acid = 0.005-0.004 = 0.001 = 10-3 moles of HCl Total volume = 25 + 50 = 75cm3 75cm3 contains10-3 moles of HCl 1000 x 10-3 : – 1000cm3 = 0.0133M pH = -log[H+(aq)] = -log(0.0133M) = 1.9 v) pH = -log[H+(aq)] = 3.4, [H+(aq)] = 10-PH = 10-3.4 [H+(aq)] = anti log(-3.4) = 3.98×10-4moldm-3
MEASUREMENT OF 𝒑𝑯 BY AN ELECTRICAL METHOD
a) USING THE HYDROGEN-HALF-CELL (ELECTRODE) To measure the pH of any solution, the H-electrode is immersed in a solution of unknown [H+] and connected to a standard half-cell of known redox potential (EƟ) to constitute a cell. The cell e.m.f is measured by a valve voltmeter and from it, the pH can be calculated as shown in the (two) examples below. The redox potential of the Hydrogen-electrode is given by the Nernst equation as follows: 𝑅𝑇 EH = E𝐻Ɵ + 2𝐹 ln[H+(aq)]2, H2(g) 2H+(aq) + 2e-, but E𝐻Ɵ = 0 𝑅𝑇 EH = 0 + lnH+(aq)]2 2𝐹 𝑅𝑇 EH = + ln[H+(aq)]2 2𝐹 Hence, using this H-electrode and comparing with another half-cell, the e.m.f of the cell could be obtained from it, and the [H+] calculated. Examples 1. If the standard Cu-electrode is connected to the H-electrode and Ecell = 0.43v. Calculate the [H+] and the pH Ɵ given that E𝐶𝑢 = 0.34v and the cell diagram is pt(s)/H2(g), 2H+(aq) ⁞ Cu2+(aq)/Cu(s) 1.0M But Ecell = ERHS – ELHS From Nernst equation Ɵ 0.06 ERHS = E𝐶𝑢 + ln[Cu2+] Ɵ 2+ = E𝐶𝑢 , ([Cu ] = 1), Ɵ 𝑅𝑇 Ecell = E𝐶𝑢 – 2𝐹 ln[H+(aq)]2 Ɵ 0.06 Ecell = (E𝐶𝑢 – ln[H+(aq)]2), [Cu2+(aq)] = 1.0M 0.43 = (0.34 – 0.34log[H+(aq)]2 0.34-0.43 log[H+(aq)]2 = = -3.0 0.03 [H+(aq)]2 = √antilog(-3.0) = 3.2 x 10-2moldm-3 =>pH = -log[H+] = 1.5 If the standard electrode is the Ag-electrode and the cell e.m.f is 1.88v, EAgƟ + = 0.8v. Find the pH and the [H+(aq)]. Cell diagram: pt(s)/H2(g), 2H2+(aq) ⁞ 2Ag+(aq)/2Ag(s) 1.0M 0.06 Ecell = EAgƟ+ – E𝐻Ɵ – 2 ln[H+(aq)]2, [Ag+(aq)] = 1.0M 0.88 = 0.8 – 0 – 0.03log[H+(aq)]2 0.8−0.88 −0.08 = log[H+(aq)]2 = = -2.667 0.03 0.03 [H+(aq)] = √antilog(-2.667) = 4.64 x 10-2 moldm-3 ∴ pH = -log[H+(aq)] = -log(4.64 x 10-2) pH = 1.33, Or log[H+(aq)]2 = 2 log[H+(aq)] = -2.667 −2.667 -log[H+(aq)] = = 1.33 ∴ pH = 1.33 However, the H-electrode has some disadvantages:- i) It is too bulky ii) ii) Slow to react to equilibrium iii) Easily poisoned by impurities Consequently, an alternative electrode that is used very conveniently is the Glass electrode b) USING THE GLASS ELECTRODE OR 𝒑𝑯 METER The glass electrode consists of silver/silver chloride electrode (a silver wire with AgCl) in a buffer solution (solution of constant pH). The electrode is placed inside a thin glass membrane permeable only to H+ ions. When the glass electrode is placed in a solution of unknown [H+] or pH, a potential difference is set up on the surface of the glass and this whole set up can be used as a half-cell. When it is combined with a reference electrode, (e.g. the calomel electrode in the cell above) a full cell is formed and the e.m.f is measured using a valve voltmeter. The calomel electrode consists of a platinum wire dipping into mercury under mercury (I) chloride in a saturated KCl solution. This arrangement is attached to a valve voltmeter calibrated to form a pH meter which gives direct readings of the pH of the solution. It is portable and can be used in the field. The cell diagram above can be given thus: – (Ag(s)/AgCl(s)/HCl(0.1M)/glass/test solution ⁞ KCl(sat)/Ag2Cl2(s)/Ag(l)) Glass electrode Calomel electrode
DISSOCIATION CONSTANTS FOR WEAK ACIDS (Ka) AND WEAK BASES (Kb) AND pH
The pH of a solution is often used to indicate the strength of an acid or base, but this is very limited since the pH of any solution changes with concentration. However, in dealing with weak acids and bases, the values of an acid (Ka) or base (Kb) dissociation or ionization constant gives a more useful means of representing the strength of the acid or base. a) ACID DISSOCITION CONSTANT Ka FOR WEAK ACIDS, SOLUTION OF WEAK ACIDS AND pH. A weak acid, HA dissociates partially thus; HA(aq) H+(aq) + A-(aq) [H+(aq)]eqm[A+(aq)]eqm Kc = Ka = [HA+(aq)]eqm + Ka[HA+(aq)] [H ] = [ A- ] [HA] log[H+] = log Ka + log [A-] [HA] -log[H+] = -log Ka – log [A-] [HA] pH = pKa – log [A-] [A-] Or pH = pKa + log [HA] Hence, the pH, pKa, [H+] and the degree of dissociation/ionization of the acid can be calculated. In the calculations it must be noted that: i) [HA]eqm = [HA]initial – [H+]eqm ii) but [H+] = [A-] (conjugate base) [H+]2 [H+]2 Ka = = [HA]eqm [HA]inital-[H+] Examples 1. A 0.1M solution of acetic acid has a pH of 2.88 at 250C. Calculate the value of its dissociation constant at this temperature. Solution pH = -log[H+] [H+] = anti(-2.88) = 1.32 x 10-3moldm-3 CH3COOH(aq) CH3COO-(aq) + H+(aq) Initial concentrations: 0.1 0 0 Equilibrium concentrations: 0.1 – 0.00132 0.00132 0.00132 [H+(aq)]eqm[CH3COO-(aq)]eqm [H+]2 Ka = [CH3COOH(aq)]eqm = [CH 3COOH] [0.00132]2 Ka = = 1.77 x 10-5moldm-3 (0.1-0.00132) 2. Calculate the pH of a 0.05M acetic acid given that its dissociation constant, Ka = 1.8 x 10-5 at 250C. CH3COOH(aq) CH3COO-(aq) + H+(aq) Initial concentrations: 0.05 0 0 Equilibrium concentrations: 0.05 – X
X X
[H (aq)]eqm[CH3COO-(aq)]eqm + [X] 2 Ka = [CH3COOH(aq)]eqm = [0.5 = 1.8 x 10-5 – X] N.B. Assumption: For weak electrolytes (acids or bases), X is very small. i.e. X<<0.05, => 0.05 – X = 0.05 x2 = 1.8 x 10-5 0.05 x = √1.8×10-5× 0.05 = 9.4 x 10-4 x = [H+] = 9.4 x 10-4 pH = log[H+] = log(9.4 x 10-4) = 3.0 [H+] pH = pKa + log [HA]
x 10-4
= -log(1.8 x 10-5) + log(0.05- 9.4 x 10-4) = 4.74 -1.71 ≈ 3.1 N.B. For every acid there is a corresponding conjugate base. A strong acid has a weak conjugate base and a weak acid has a strong conjugate base, thus acetate ions, CH3COO- from acetic acid are a stronger base than Cl- ions from HCl acid. The basicity of an acid is the number of replaceable protons which it can loose per molecule of the acid, e.g. H2SO4 is dibase or diprotic, H3PO4 is tribasic or tripotic, etc. 3. Calculate Ka, degree of ionization and percentage of ionization of propanoic acid HC3H5O2. The pH of a 0.012M aqueous solution in 3.40. Let HC3H5O2 be HPIC HPIC (aq) H+(aq) + PIC-(aq) Initial concentrations: 0.012 0 0 Equilibrium concentrations: 0.012 – x X X [H+][PIC-] X2 Ka = = [HPIC] 0.012-X Since pH = -log[H+] = 3.4 [H+] = anti log(-3.4) = 3.98 x 10-4moldm-3 = X X = 0.000398moldm-3 << 0.012 X2 (0.000398)2 Ka = = = 1.32 x 10-5moldm-3 0.012-X 0.012 X = 0.00398mole out of every 0.012mole acid that ionizes. 0.00398 ∴ degree of ionization = = 0.033. 0.012 % of ionization = degree of ionization x 100 = 0.033 x 100 = 3.3%
Indicators
Indicators are substances which change colour according to the [H+] of the solution to which they are added. Acid-base indicators are weak acids or weak bases whose acid or base conjugate are coloured different. They are used in: – i) Testing acidity and alkalinity and thus the pH of solutions ii) Determining the end- point in acid-base titrations
Functioning Of An Indicator
Let Hln be an indicator which is a weak acid and it undergoes partial dissociation thus Hln(aq) H+(aq) + ln-(aq) Acid Conjugate base Colour A Colour B In an acid medium where [H+] is high, the equilibrium is displaced to the left favoring the production of more Hln molecules thus the intensity of colour A increases. The addition of OH- or NH3 removes the H+ ions to form H2O or NH4+, thus displacing the equilibrium to the right to produce more ln- and the intensity of colour B increases. The equilibrium constant for the dissociation of the indicator is thus [H+][ln-] Kin = [Hln] [𝐻𝑙𝑛] [H+] = Kin – [ln ] ln-] pH = pKin + log [𝐻𝑙𝑛] [ln-] The colour observed depends on the ratio and on the [H+]. The [Hln] and [ln-] in solution depends on [𝐻𝑙𝑛] the [H+] in solution so that the change of colour of an indicator depends on certain range of [H+] or pH, e.g. pH of phenolphthalein is 8-10. Hence, the appearance of the coloration starts at 8 and is completely developed at 10. The rate of change of pH is related to the rate of change of indicator colour. Towards the end- point of a titration the indicator changes colour sharply on addition of alkali. At the exact end- point of the titration, the indicator colour is mid-way between acid (Hln) and alkali (ln-) and the (Hln) = (ln-), then pH = pKin. For the colour change of the indicator to be complete, let’s assume that the colour change from (Hln) colour to that of (ln-) will be complete when (ln-) = 10(Hln) (ln-) i.e. = 10, colour B is observed. (Hln) Similarly, when (Hln) = 10(ln-) (Hln) i.e = 10, colour A is observed. (ln-) (Hln) When (ln-) = 10(Hln), = 10, (ln-) [H+][ln-] Kin Kin = = 10[H+], [H+] = [Hln] 10 [Hln] And when [Hln] = 10l/n-], = 10, [ln-] [H+] Kin = , [H+] = 10Kin. Most indicators are sensitive to the eye, hence good for titrations because they have a pH range of only two units. The pH range of an indicator should be at the point where there is the greatest variation in pH. COMMON ACID-BASE INDICATORS INDICATOR ACID COLOUR BASE COLOUR pH RANGE Phenolphthalein Colourless Red/pink 8-10 Litmus Red Blue 6-8 Methyl red Pink Yellow 4-6 Methyl orange Pink Orange 3-5 N.B A non-acid-base indicator is starch, used in iodometric titrations VARIATION OF pH DURING ACID-BASE TITRATIONS: An acid- base titration is a technique used to measure the volumes of acid and base required for complete neutralization. The neutralization of an acid by a base is simply given as follows: H+(aq) + OH-(aq) H2O(l). From this equation, the [H+] or pH is expected to vary during acid-base titration. The variation depends on the strength of the acid and that of the base, but at each point, Kw = [H+][OH-] = 10-14. A pH meter can be used to follow the variation and to indicate the end- point of the reaction. A titration curve can be used to choose a suitable indicator which will show the end-point of the titration. Base in the conical flask and acid in the burette SUMMARY OF TITTRATIONS, pH RANGES AND INDICATORS No TITRATION MARKED pH RANGE INDICATORS 1 Strong acid-strong base( alkali) 4-10 Any indicator 2 Strong acid-weak base 3.5-6.5 Methyl red or methyl orange 3 Weak acid-strong base 7.5-10.5 Phenolphthalein 4 Weak acid-weak base No marked change End-point cannot be detected .
Titration Curves
a) STRONGACID- STRONG BASE. E.g. Titration of 25cm3 of 0.1MHCl with 0.1M NaOH. All common indicators change colour within this range, and so the end- point (when indicator changes colour) coincides with the equivalence point. The titration causes a large 𝑝𝐻 change on neutralization as shown on the graph, because both the acid and base are fully ionized. The equivalent point of this titration occurs when the amount of acid and base are exactly equal. b) STRONG ACID-WEAK BASE. E.g. 25cm3 of 0.1MHCl in 0.1MNH3 The 𝑝𝐻 at the end- point falls in the acid zone (𝑝𝐻 < 7). Hence, indicators of 𝑝𝐻 within acid zone are suitable, e.g. methyl orange or methyl red. c) WEAK ACID-STRONG BASE: E.g. 25cm3 of 0.1M acetic acid and 0.1MNaOH The end- point falls in the alkaline zone (pH > 7). Hence indicators whose pH ranges are above 7 are suitable e.g. phenolphthalein d) WEAK ACID-WEAK BASE: No suitable indicator because of no marked change in pH. e.g. CH3COOH(aq) and NH3(aq) VARIATION OF CONDUCTIVITY DURING TITRATION (WEAK ACID-WEAK BASE) H+ ions are smaller in size than OH- ions and can thus move faster than OH-.This explains why H+ ions have a higher conductivity than OH- ions. During titration of a weak acid with a weak base in the burette, the [H+] decreases and so conductivity decreases. It passes through a minimum when the concentration of the ions in the solution are minimum, i.e. [H+] = [OH-] = 10-7moldm-3, i.e. equivalent point. Hence, VEP(cm3) of base neutralize 25cm3 of acid thus the molarity of the base can be calculated. If more base is added, the OH- carry the conductivity leading to a gradual rise in conductivity.
Salt Hydrolysis
The equivalence point in titration does not always correspond with the production of a neutral solution, i.e. pH= 7.0. Only pure or distilled water, and aqueous solutions of salts of strong acid and strong base will have a pH of 7 or be neutral. Solutions of other salts are either acidic or alkaline. Salts of weak acids and strong bases are alkaline, e.g. CH3COONa; those of strong acids and weak bases are acidic e.g. NH4Cl. Those of weak acids and weak bases are either acidic or alkaline depending on the Ka and Kb of the acid and base respectively. The acidity or alkalinity of these salts is due to hydrolysis of the base conjugate of the weak acid salt or hydrolysis of the acid conjugate of the weak base salt. DEFINITION: HYDROLYSIS – Is the reaction of ions from an acid or base salt with water to form the undissociated weak acid or weak base. a) HYDROLYSIS OF A SALT OF WEAK ACID/STRONG BASE. E.g. CH3CO2Na CH3CO2Na(aq) CH3COO-(aq) + Na+(aq) ———–(i) H2O(l) H+(aq) + OH-(aq) ————————————-(ii) CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq) ———-(iii) The interaction of CH3COO- ions with water [equation (iii)] removes H+ from solution leaving the solution with an excess of OH- ions so that the pH is greater than 7. Therefore, the end -point for the NaOH + CH3CO2H titration is greater than 7. b) HYDROLYSIS OF A SALT OF A STRONG ACID-WEAK BASE, E.g. NH4Cl NH4Cl(aq) NH4+(aq) +Cl-(aq) ———————–(i) H2O(aq) H+(aq) + OH-(aq) ———————-(ii) NH4+(aq) + H2O(l) NH4OH-(aq) + H+(aq) ——————-(iii) Equation (iii) makes the [H+] > [OH-]. Therefore, the pH < 7, i.e. for the titration. HCl + NH3, end- point has pH < 7.
Buffer Solutions
DEFINITION: A buffer solution is one that resists changes in PH on addition of a small amount of a strong acid or strong base. A buffer solution consists of either a weak acid and its salt with a strong base, e.g., CH3COOH/CH2COONa, or a weak base and its salt with a strong acid, e.g., NH3/NH4Cl, etc. A buffer solution maintains a constant pH. a) An acid buffer solution comprises a mixture of a weak acid and its salt with a strong base, e.g. acetic acid (CH3COOH) and sodium acetate (CH3COONa) and maintains a nearly constant pH value < 7. CH3CO2H(aq) CH3COO-(aq) + H+(aq) (Ka) CH3COONa(aq) CH3COO-(aq) + Na+(aq) (Kb) H2O(l) H+(aq) + OH-(aq) (Kw) Other examples: H2CO3(aq)/Na2CO3(aq), H3PO4/K3PO4(aq), C6H5COOH(aq)/C6H5COONa(aq), HF/NaF. H2CO3/NaHCO3 is the buffer in human blood which maintains blood PH at 7.4.A PH value >7.8 or < 6.8 can lead to an abrupt death in humans. b) An alkali (basic) buffer solution comprises a weak base and its salt with a strong acid, e.g. ammonia and ammonium chloride, with a nearly constant pH value>, NH3/(NH4)2SO4, NH3/NH4NO3. NH4Cl(aq) NH4+(aq) + Cl-(aq) NH4+(aq) NH3(aq) + H+(aq) H2O(l) H+(aq) + OH-(aq)
Effects Of Adding Small Amounts Of Strong Acid Or Base To A Buffer
Solution
Any buffer solution is characterized by the ability to neutralize either added acid or base and maintain its pH. a) TO AN ACID BUFFER, e.g. CH3COOH/CH3COONa. Consider the equations H2O(l) H+(aq) + OH-(aq) ——————————————(i) (Kw) CH3COOH(aq) CH3COO-(aq) + H+(aq) —————————-(ii) (Ka) CH3COONa(aq) CH3COO-(aq) + Na+(aq) ————–(iii) i) ADDITION OF ACID: e.g. 10cm3 of HCl acid to 1dm3 of the buffer given that they have the same molarities’ HCl (aq) H+(aq) + Cl-(aq) The [H+] increases thereby interfering with equilibrium (ii). The added H+ ions are thus used up to form CH3COOH acid according to Le Chatelier’s principle, thus restoring the Ka and as such, the [H +] remains constant in solution. Consequently, the pH remains unchanged. In practice, the pH changes very slightly (negligible). ii) ADDITION OF STRONG BASE, e.g. NaOH NaOH(aq) Na+(aq) + OH-(aq) The added OH- ions react with H+ ions to form water with equation (ii). Hence, the [H+] drops and more acid ionizes to maintain Ka in equation (ii). The dissociation restores the [H+] and the pH remains unchanged. b) TO AN ACID BUFFER, e.g. NH3/NH4Cl Consider the equations: NH4Cl(aq) NH4+(aq) +Cl-(aq) ————————————-(i) NH4+(aq) NH3(aq) + H+(aq) ————————————————-(ii) H2O(l) H+(aq) + OH-(aq) —————————————————(iii) i) ADDITION OF A STRONG ACID, e.g. HCl HCl(aq) H+(aq) + Cl-(aq) This increases the [H+] which reacts with equation (ii) to form the acid, restore Ka, [H+] and pH ii) ADDITION OF A STRONG BASE, e.g. NaOH NaOH(aq) Na+(aq) + OH-(aq) The additional OH- reacts with H+ to give H2O by equation (ii) or with the acid to give base equation (iv). This decreases the [H+] and affects equation (ii) so that the NH4+ dissociates to restore [H+], Ka and pH.
CALCULATION OF THE pH OF A BUFFER SOLUTION (pH CALCULATIONS OF
Buffers)
For the acid buffer, the weak acid is HA, Ha(aq) H+(aq) +A-(aq) [H+]eqm[H-]eqm 𝐾𝑎[HA] Ka = [H+] = [HA]eqm [A-] [H+] pH = pKa + log [HA] The pH depends on the ratio of the concentration of the acid and base but not on the actual amounts at a given temperature. The acid in the buffer always has a very small Ka so that the [base], [A-] is always taken to have resulted from the salt dissociation. Also, [acid] eqm = [undissociated acid] = [acid initial]. Hence, for an acid buffer [base] pH = pKa + log [acid] [salt] Or pH = pKa + log [acid] N.B. For an alkaline buffer, similar deduction on the equation. B(aq) + H2O(l) BH+(aq) + OH-(aq) [BH+][OH-] Kb = = [𝐵] [salt] pOH = pKb + log [acid] But pH + pOH = 14 pH = 14 – pOH
Uses Of Buffers
1. Buffer solutions are used in analytic chemistry and Biochemistry, e.g. in the preparation of solutions of known pH for the calibration of pH meters. 2. In medicine intravenous injections and most drugs are buffered to maintain blood pH to the normal value of 7.4 ( H2CO3/NaHCO3, buffer in human blood). 3. Fermentation processes have to be buffered otherwise pH changes may kill the fermenting bacteria, e.g, production of alcoholic drinks. 4. Buffers are equally useful in maintaining the pH of the soil, tin food and milk.
Assignment
1. 0.83g of CH3COONa were dissolved in 500ml of 0.5M CH3COOH, if the pH of the solution was 3.3. Calculate the Ka for CH3COOH. If the Ka for CH3COOH is 1.8 x 10-5. Calculate the pH of a solution containing 0.06M CH3CO2H and 0.2M CH3CO2Na. 2. What is a buffer? Give two examples. Explain with the aid of equations the reaction that takes place in each buffer solution given as an example upon addition of small amount of acid and alkali.
Solution
1. 500ml dissolve 0.83g of CH3COONa 1000 x 08.3 : – 100ml = 1.66g Molar mass of CH3COONa = 2(12) + 3(1) + 2(16) + 23 = 82 mass per dm3 1.66 : – [CH3COONa] = = = 0.02M Molar mass 82 [salt] [salt] 0.02 pH = pKa + log pKa = pH – log = 3.3 – log( ) = 3.26 [acid] [acid] 0.5 Ka = antilog(-3.26) = 5.495 x 10-4moldm-3 [salt] pH = pKa + log [acid] 0.2 = -10log(1.8 x 10-5) + ( ) 0.06 = 4.74 + 0.52 = 5.26 2. If the Ka of acetic acid is 1.8 x 10-5mold-3, calculate the pH of a mixture of a). 0.1M acetic acid and 0.2M sodium acetate. b)i) Calculate the new pH when 10cm3 of 1M HCl is added ii) 10cm3 of 1.0NaOH is added c) Comment on the results.
Solution
[salt] a) pH = pKa + log [acid] 0.2 = -log(1.8 x 10-5) + log 0.1 = 4.75 + 0.30 = 5.04 b)i) Number of moles of HCl acid = x 1.0 = 0.01 HCl(aq) H+(aq) + Cl-(aq) CH3COOH(aq) CH3COO-(aq) + H+(aq) 0.1 0.1 CH3COONa(aq) CH3COO-(aq) + Na+(aq) 0.2 0.2 : – New [CH3COOH] = 0.1 + 0.1 = 0.11M : – New [CH3COONa] = 0.2 – 0.1 = 0.09M 0.19 : – New pH = 4.74+ log0.11 = 4.98 ii) Number of moles of OH- added = x 1.0 = 0.01 : – New [CH3COONH = 0.1 – 0.01 = 0.09M : – New [CH3COONa] = 0.2 – 0.1 = 0.21M 0.21 : – New pH = 4.74 + log0.09 = 5.11 The change in pH in each case is very slight (negligible)
Assignment
Exercise: Q. 4, P. 156, Q. 11, 14, 10, 16. MCQs: Q. 1- 40, P. 147 – 149
Solubility Equilibria
1. Solubility Product (Ksp) DEFINITION: Solubility product ( Ksp) of a salt is the product of the concentrations of the ions of the salt in a saturated solution of the salt raised to the powers of their respective coefficients in the dissociation equation at a given temperature. Consider an electrolyte, AxBy(s), which dissociates partially in water as follows AxBy(s) xAy+(aq) + yBx-(aq) x y [Ay+(aq)] eqm [Bx-(aq)]eqm Equilibrium constant, Kc = [AxBy(s)(aq)]eqm 𝑥 𝑦 Kc[AxBy(s)(aq)]eqm = [Ay+(aq)]𝑒𝑞𝑚[ Bx-(aq)]𝑒𝑞𝑚 N.B. Ksp applies only to sparingly soluble salts with a total concentration less than 0.01M. EXAMPLES OF Ksp EXPRESSIONS 1. AgCl(s) Ag+(aq) + Cl-(aq) Ksp = [Ag+(aq)]eqm[Cl-(aq)]eqm 2. PbCl2(s) Pb2+(aq) + 2Cl-(aq) Ksp = [Pb2+(aq)]eqm[Cl-(aq)]2eqm 3. Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq) Ksp = [Ca2+(aq)]3eqm[PO43-(aq)]2eqm If the solubility of a sparingly soluble salt is known, its Ksp can be calculated and vice versa. Also, if the concentration of one of the ions is known, the Ksp can equally be calculated.
Examples
It is experimentally found that 1.2 x 10-3moles of PbI2 is dissolved in 1dm3 of aqueous solution at 250C. What is the solubility product constant at this temperature? PbI2(s) Pb2+(aq) + 2I-(aq) Eqm concentration: 1.2 x 10-3moldm-3, 2(1.2 x 10-3moldm-3) Ksp = [Pb2+]eqm[I-]2eqm = (1.2 x 10-3moldm-3)( 1.2 x 10-3moldm-3)2 = 6.9 x 10-9moldm-9 2. The solubility of the Ag2CO3 is 1.16 x 10-4moldm-3 at 200C. Calculate the Ksp of Ag2CO3 at this temperature. Ag2CO3(s) Ag2+(aq) + CO32-(aq) m n Eq conct : S 2S S Ksp = [Ag2+]2eqm[ CO32-]eqm = (2S)2(S) = 4S3 = 4(1.16 x 10-4)3 = 6.24 x 10-12moldm-9 2. COMMON ION EFFECT AND SOLUBILITY Ksp is an equilibrium constant and if applied to a saturated solution of a salt, MA(s), MA(s) M2+(aq) + A2-(aq), to which is added a solution containing M2+ ions, the equilibrium moves to the left to remove the excess M2+ ions, maintaining the Ksp (by Le Chatelier’s principle). Thus the solubility of MA(s) reduces. DEFINITION: common ion effect is the precipitation of a solute in solution on addition of an electrolyte solution which has an ion in common with the solute.
Examples
The solubility product, Ksp of BaSO4 is 1.0 x 10-10mol2md-6. Calculate the solubility of BaSO4 in a) water, b) 0.1MNa2SO4 a) BaSO4(s) Ba2+(aq) + SO42-(aq) Eqm. Conctn. a a a : – Ksp = [Ba2+]eqm[SO42-]eqm = (a)(a) = a2 = 1.0 x 10-10mol2md-6 a = √1.0 x 10-10 = 10-5moldm-3 b) Let the new solubility be Smoldm-3 in Na2SO4, BaSO4(s) Ba2+(aq) + SO42-(aq) Eqm. conctn. S S S Na2SO4(aq) 2 Na (aq) + SO42-(aq) + : – [SO42-] = S + 0.1 [Ba2+] = S : – Ksp = [Ba2+]eqm[SO42-]eqm = (S)( S + 0.1) = 1.0 x 10-10 = S(S +0.1) = 1.0 x 10-10 But S << 0.1 (or use quadratic formulae to solve the equation). S + 0.1 = 0.1 : – S(0.1) = 1.0 x 10-10 1.0 x 10-10 S= = 1.0 x 10-9moldm-3 0.1 Hence, solubility in H2O is far greater than in Na2SO4 solution, because of the common ion effect.
Assignment
What is the molar solubility of lead chloride, PbCl2, in 0.08M sodium chloride solution? The solubility product constant for PbCl2 is 1.6 x 10-5. PbCl2(s) Pb2+(aq) + 2Cl-(aq) Eqm. conctn. S S 2S NaCl(aq) Na+(aq) + Cl-(aq) 0.08 0.08 [Cl-] = 2S + 0.08 [Pb2+] = S Ksp = [Pb2+]eqm[Cl-]2eqm = (S)(2S + 0.08) = 1.6 x 10-5 Let’s assume that 2S << 0.08 Ksp = S(2S + 0.08)2 = 1.6 x 10-5 = S(0.08)2 = 1.6 x 10-5 1.6 x 10-5 S= = 2.50 x 10-3M S(0.08)2 The molar solubility of PbCl2 in 0.08MNaCl is 2.50 x 10-3M 3. APPLICATIONS OF SOLUBILITY PRODUCT 1. In qualitative analysis – The precipitation of insoluble salts is used in identifying anions, e.g. halides. 2. Precipitation titration: – The Ksp determines the maximum concentration of ions in solution at a given temperature. Chloride solutions are estimated by titrating against AgNO3 solution with potassium chromate as indicator. Potassium chromate gives a red precipitate of Ag2CrO4 at end- point during titration. Ag2CrO4 should not precipitate until all chloride ions have precipitated as AgCl. At the end- point the concentration of (Cl-) can be found. Ksp(AgCl) = 1.2 x 10-10mol2dm-6 Ksp(Ag2CrO4) = 2.4 x 10-12mol3dm-9 Ag2CrO4 is more soluble than AgCl AgCl(s) Ag+(aq) + Cl-(aq)
S S S
Then Ksp = [Ag+]eqm[Cl-]eqm = [Ag+]2 = 1.2 x 10-10mol2dm-6 [Ag+] = 1.1 x 10-5moldm-3 Ag2CrO4(s) 2Ag+(aq) + CrO42-(aq) S 2S S Ksp = [Ag+]2eqm[CrO42-]eqm = 2.4 x 10-12 2.4 x 10-12 2.4 x 10-12 [CrO42-] = = [2(1.1 x 10-5)]2 4.84 x 10-10 = 4.98 x 10-3moldm-3 N.B. K2CrO4 is only used as an indicator in an alkaline/neutral medium because the red Ag2CrO4 precipitate formed at end -point is soluble in an acid medium. Hence, to titrate HCl acid against AgNO3 using K2CrO4 as an indicator, the acid is first neutralized using CaCO3. Excess CaCO3 gives an alkaline medium, still good for the titration.
