Presentation on theme: "PH, K a, pK a, K w pH of strong and weak acid and strong base Acid-base indicators Change in pH of acid-base titration Buffer solution Solubility product,"— Presentation transcript:
pH, K a, pK a, K w pH of strong and weak acid and strong base Acid-base indicators Change in pH of acid-base titration Buffer solution Solubility product, K sp Common ion effect Ionic Equilibria
The pH scale pH is a measure of [H + (aq)] in an aqueous solution. pH = -log 10 [H + ] In most aqueous solution, the pH is between the range of 0 – 14. pH values depend on concentration of acid or base and degree of dissociation, α. pH of an acid increases upon dilution at constant temperature to a volume V.
pH of a base decreases upon dilution at constant temperature to a volume V. Therefore, using the dissociation constant to measure pH is more reliable because it is a constant at all dilution and only will be influenced by temperature.
Ionic product of water Water is slightly ionised : H 2 O H + + OH - H = positive K c = [H + ]. [OH - ] [H 2 O] Degree of ionisation of water is very small, hence [H 2 O] remains virtually constant. [H + ].[OH - ] = K c x [H 2 O] = constant = K w K w = [H + ]. [OH - ] K w is the ionic product of water. K w depend on temperature. At 25 C, K w = 1.0 x mol 2 dm -6. In pure water, 25 C : [H + ] = [OH - ] Hence [H + ] = [OH - ] = (1.0 x ) 1/2 = 1.0 x mol dm -3 Therefore, pH of water = -log 10 (1.0 x ) = 7.0
The relationship between pH, [H + ] and [OH - ] : K w varies with temperature. When temperature increases, K w increase. Will the pH of water be affected when temperature is changed? E.g : Calculate the pH of water at the temperature below, using the K w values given. a) At 10 C, K w = 0.1 x mol 2 dm -6. b) At 100 C, K w = 51 x mol 2 dm -6.
Sometimes the term pOH is used. pOH = -log 10 [OH - ] At 25 C, pH + pOH = 14 pOH = 14 - pH
Calculate the pH of strong acid and base Example: Calculate the pH of the following solutions mol dm -3 HCl mol dm -3 H 2 SO 4 1. A solution where [H + ] ions = 3.5 x mol dm -3 The pH of a solution of HCl is Calculate the concentration of hydrogen ions.
Calculate the pH of strong acid and base
pH of Weak Acids and Bases Acid Dissociation Constant, K a Weak Acid dissociates partially in water : HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) The acid dissociation constant, K a : K a = [H 3 O + ] [A - ] mol dm -3 [HA] pK a = -log10 K a The K a value is the measure of the strength of acids. Larger K a value (smaller pK a ), stronger acid.
Base Dissociation Constant, K b Weak base dissociates partially in water : B(aq) + H 2 O(l) BH + (aq) + OH - (aq) The base dissociation constant, K b : K a = [BH + ] [OH - ] mol dm -3 [B] pK b = -log10 K b The K b value is the measure of the strength of base. Larger K b value (smaller pK b ), stronger base.
Calculating the pH of a weak acid E.g 1: What is the pH of mol dm -3 ethanoic acid ? K a = 1.74 x mol dm -3.
E.g 2 : What is the pH of 0.05 mol dm -3 methanoic acid if its pK a is 3.75?
E.g 3 : The pH of a weak acid, HA, of concentration 0.1 mol dm -3 was found to be Calculate the value of pK a for the acid.
Exercise : 1. K a for a weak monobasic acid = 1.0 x mol dm -3. In a 0.1 mol dm - 3 solution of the acid, calculate a) the concentration of H + (aq) ions; b) the pH; c) the concentration of OH - (aq) ions. [K w = 1.0 x mol 2 dm -6 ]
2. A 0.1 mol dm -3 solution of a weak monobasic acid has a pH of 4.0. Calculate K a and pK a for the acid.
3. The acid dissociation constant, K a, for methanoic acid is 1.8 x mol dm -3. In a 0.1 mol dm -3 solution of methanoic acid, calculate: a) the concentration of hydrogen ions; b) the pH; c) the concentration of hydroxide ions; d) the degree of ionisation of then acid.
Calculating the pH of a weak base E.g 1 : What is the pH of mol dm -3 NH 3 (aq) if K a for the NH 4 + ion is 5.62 x mol dm -3 ? K w = 1.00 x mol 2 dm -6.
E.g 2 :Sodium ethanoate solution is alkaline because the ethanoate ion is a weak base and reacts with water according to the equation : CH 3 CHOO - (aq) + H 2 O(l) CH 3 COOH(aq) + OH - (aq) a) pKa for ethanoic acid is Calculate a value for K a, including the units. b) Write an expression for K a for ethanoic acid, and rearrange it to give an expression for [H + ]. c) K w is 1.00 x mol 2 dm -6. Write an expression for K w and rearrange it to give an expression for [OH - ] d) Write an expression for the concentration of CH 3 COOH at equilibrium. e) What is the equilibrium concentration of CH 3 COO - ions in 0.2 mol dm -3 sodium ethanoate solution? State any assumption you are making. f) By combining your answers to (b), (d), and (e), calculate the pH of 0.2 mol dm -3 sodium ethanoate.
Exercise : Calculate the pHs of the following solutions of weak bases. In each case. K w = 1.00 x (a) 0.05 mol dm -3 1-aminopropane solution, C 3 H 7 NH 2. pKa for C 3 H 7 NH 3 + = C 3 H 7 NH 2 + H 2 O C 3 H 7 NH OH - (b) mol dm-3 phenylamine solution, C 6 H 5 NH 2. pKa for C 6 H 5 NH 3 + = 4.62 C 6 H 5 NH 2 + H 2 O C 6 H 5 NH OH -
Buffer Solution *Is a solution whose pH remains almost unchanged when small amounts of acid or base are added to it. 2 types of buffer solution: 1) Acidic buffer mixture of weak acid with sodium salt of weak acid. E.g, CH 3 COOH and CH 3 COO - Na +. 2) Alkaline buffer mixture of a weak base with the salt of the weak base. E.g, NH 3 and NH 4 Cl. Buffer works by removing extra acid or alkali added.
Acidic Buffer CH 3 COOH CH 3 COO - + H + CH 3 COO - Na + CH 3 COO - + Na + (Salt is fully ionised, high [CH 3 COO - ]) Weak acid will only be slightly dissociated hence concentration of un- ionised CH 3 COOH is relatively high. When small amount of acid added to buffer: H + + CH 3 COO - CH 3 COOH (added) (from salt) The H + ions added combine with CH 3 COO - ion from the salt to form CH 3 COOH (additional H + ions removed). Hence pH remains almost unchanged. [salt] in buffer decrease & [acid] increase is equal to the amount of H + added.
When small amount of base is added to buffer : OH - + CH 3 COOH CH 3 COO - + H 2 O (added) The extra OH - ions added removed by reacting with the large concentration of CH 3 COOH. Hence pH remains almost unchanged. [acid] in buffer decrease and [salt] increase is equal to the amount of OH - added.
Alkaline Buffers NH 3 + H 2 O NH OH - (weak base partially ionised, concentration of NH 3 is relatively high) NH 4 Cl NH Cl - (salt is fully ionised) When small amount of acid added to buffer: H + + NH 3 NH 4 + (added) The extra H + ions added removed by reacting with the large concentration of NH 3 in the buffer. Hence pH remains almost unchanged. [base] decrease and [salt increase is equal to the amount of H + added.
When small amount of base is added to buffer : OH - + NH 4 + NH 3 + H 2 O (added) (from salt) The OH - ions added combine with NH 4 + ion from the salt to form NH 3 (additional OH - ions removed). Hence pH remains almost unchanged. [salt] decrease and [NH 3 ] increase is equal to the amount of OH - added.
Application of Buffer Solution 1) HCO 3 - ion act as the principle buffer in human blood. When acid enters the blood, the extra H 3 O + ions are removed by HCO 3 - ions which act as a base: HCO H 3 O + H 2 CO 3 + H 2 O Or HCO H + H 2 CO 3 Or HCO H + H 2 O + CO 2 When base enters the blood, extra OH- ion are removed by carbonic acid. H 2 CO 3 + 2OH - CO H 2 O
2) H 2 PO 4 - and HPO 4 2- ions also act as a buffer in blood or saliva. H 2 PO 4 - ion is a weak acid and dissociates partially in aqueous solution : H 2 PO 4 - HPO H + When acid is added, the equilibrium shift to the left, extra H + is consumed, pH maintained. [H 2 PO 4 - ] increases. HPO H + H 2 PO 4 - When alkali is added, equilibrium shift to the right because the OH - ion added will react with the H + ions in the buffer. OH - + H + H 2 O H 2 PO OH - HPO H 2 O OH - ions removed, pH maintained.
3) Other usage of buffer solution : Solution of known pH for checking instruments and indicators. Biochemistry experiments.
Calculation of pH of buffer solution In an acidic buffer : HA H + + A - (partial dissociation) MA M + + A - K a = [H + ]. [A - ] [HA] Where [HA] = concentration of HA acid in the mixture. [A - ] = concentration of A - from MA (the amount of A - from HA is very small, can be ignored) [H + ] = K a x [HA] [A - ]
[H + ] K a x [acid] or pH = pK a + log [salt] [salt] [acid] This equation explains why the pH is affected very little by dilution because the ratio of [acid]/[salt] remains constant on dilution. [H + ] K a Example 1 : Acidic buffer A buffer solution made by mixing 3.28 g dm -3 of CH 3 COO - Na + and 0.01 mol dm -3 CH 3 COOH. [K a (CH 3 COOH) = 1.7 x mol dm -3 ] (a)What is the pH of the buffer.
(b) Calculate the change in pH when 1 cm 3 of 1.0 mol dm -3 NaOH is added to 1 dm 3 of buffer. (c) Calculate the change in pH when 1 cm 3 of 1.0 mol dm -3 NaOH is added to 1 dm 3 of 0.01 mol dm -3 CH 3 COOH.
Example 2 : Alkaline buffer. A buffer solution contains 1.00 mol dm -3 NH 3 and 0.40 mol dm -3 NH 4 Cl. [K a (NH 4 + ) = 5.62 x mol dm -3 ] (a) Calculate the pH of the buffer. (b) Calculate the effect on the pH when 5.00 cm 3 of 10.0 mol dm -3 HCl solution is added to 1000cm 3 of the buffer solution.
Exercise: In what proportions should ammonia and ammonium chloride be mixed in solution to give a buffer solution of pH 10.0? pK a (NH 4 + ) = 9.25.
Hydrolysis of Salts Salt ofSolution formed in water Reaction (example) Weak acid & strong base Alkaline solutionCH 3 COONa(aq) CH 3 COO - (aq) + Na + (aq) -CH 3 COO - ion reacts with H + ions from water. CH 3 COO - + H + (aq) CH 3 COOH(aq) -Most H + removed because CH 3 COOH is a weak acid. [H + ] decrease, pH > 7. - [OH - ] > [H + ] Strong acid & weak base Acidic solution(NH 4 ) 2 SO 4 (aq) 2NH 4 + (aq) + SO 4 2- (aq) - NH 4 + ions react with OH - ions from water. NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O(l) -Most OH - removed because NH 3 is a weak base. -As OH - ions are removed, more water ionises to keep K w constant. [H + ] increases, pH < 7 - [H + ] > [OH - ] In water, [H + ] = [OH - ] Neutral
Hydrolysis of Salts Salt ofSolution formed in water Reaction (example) Weak acid & weak base Neutral solutionCH 3 COONH 4 (aq) CH 3 COO - (aq) + NH 4 + (aq) -CH 3 COO - ions react with H + from water to form CH 3 COOH. NH 4 + ion react with OH - ions from water to form NH 3 & H 2 O. -Both H + and OH - are equally removed, pH remains at 7. - [H + ] = [OH - ] Strong acid & strong base Neutral solutionNaCl Na + + Cl - -NaCl just dissolves in water hence concentration of H + & OH - remains. [H + ] = [OH - ]
Acid-base Indicators Acid-base indicators are substances which change colour according to the hydrogen ions concentration of the solution to which they are added. Most indicators are weak acids. They have an acid colour (unionised molecule, HIn) and the alkaline colour (anion, In - ). Each indicator has a pH range over which it changes colour. Example:
How an indicator works. Indicators can be regarded as weak acids. HIn(aq) H + (aq) + In - (aq) K a = KIn = [H + ]. [In - ] [HIn] At the end point of titration, [HIn] = [In - ] KIn = [H + ] pKIn = pH At the end point, both acid colour and alkaline colour form will be present in appreciable quantities ([HIn] = [In - ]) It is not possible to determine precisely when the two forms are at equal concentration. Hence indicators change colour over a range of about 2 pH units.
In titration, an acid-base indicator is used to mark the end-point. For an accurate result : a) The indicator must change colour sharply on additon of 1 or 2 drops of liquid from the burette. b) The colour change must occur when the correct volume of liquid is added from the burette.
pH change during titrations 1. Strong acid + strong base titration. 50cm 3 of 0.1 mol dm -3 NaOH + 25 cm 3 of 0.1 mol dm -3 HCl.
pH curve begins at pH 1, indicating the presence of strong acid. pH curve ends at pH 13, indicating presence of strong base. As NaOH is added, the pH increases slowly, but increases rapidly between the range of 3 – 11. The sharp increase in pH indicates the end point of titration. The end-point is the mid-point of the vertical line of the titration curve (pH = 7). Almost any indicator can be used but bromothymol blue is the ideal indicator.
2. Strong acid + weak base 50cm 3 of 0.1 mol dm -3 NH cm 3 of 0.1 mol dm -3 HCl.
pH curve begins at pH 1, indicating the presence of strong acid. The graph ends at pH 11, showing presence of weak base. As NH 3 is added, theres a slow increase in pH, but increases rapidly between the range of 3 – 7. The end-point of the solution is acidic (pH 5.5). This is due to hydrolysis of NH 4 Cl salt. The vertical section of the curve is shorter than titration of strong acid and strong base. Methyl orange is the ideal indicator. Phenolphthalein cannot be used because it will change colour at the wrong volume of NH 3 solution (about 30 cm 3 ).
3. Weak acid + strong base 50cm 3 of 0.1 mol dm -3 NaOH + 25 cm 3 of 0.1 mol dm -3 CH 3 COOH.
pH curve begins at pH 3, indicating the presence of weak acid. The graph ends at pH 13, showing presence of strong base. When half the acid is neutralised, [CH 3 COOH] = [CH 3 COONa] max. buffer capacity This is a buffer solution and the pH only changes gradually at the buffer region. The end-point of the solution is alkaline (pH 8.5). This is due to hydrolysis of CH 3 COONa salt. The vertical section of the curve is shorter than titration of strong acid and strong base. Phenolphthalein is the ideal indicator. Methyl orange is unsuitable because it will change colour very slowly over a large volume of NaOH.
The buffer region corresponds to pH = pK a ± 1. When : [acid] = 10, pH = pK a -1 [salt] 1 [acid] = 1, pH = pK a [salt] 1 [acid] = 1, pH = pK a +1 [salt] 10 Acidic buffer is most effective (at max. buffer capacity) when [acid]=[salt], pH = pK a
Alkaline buffer is most effective (at max. buffer capacity) when [base]=[salt], pOH = pK b In the buffer region, the pH is insensitive to small changes in concentration of acid or base.
4. Weak acid + weak base 50cm 3 of 0.1 mol dm -3 NH cm 3 of 0.1 mol dm -3 CH 3 COOH.
pH curve begins at pH 3, indicating the presence of weak acid. The graph ends at pH 11, showing presence of weak base. At the end point, the solution (CH 3 COONH 4 ) is approximately neutral. Although the salt is hydrolysed, both H + and OH - from water are removed equally. No net change in pH. There is no straight vertical section on the graph. No indicator can be used to detect the end-point as the colour change is always gradual. A pH meter is used during titration potentiaometric titration. To find the end-point, a graph of pH vs. volume must be drawn, and the end-point (at pH 7) determined from the graph.
How to find end point pH and volume when the graph has no vertical section?
5. Titration of polybasic acids E.g 1 : 0.1 mol dm -3 NaOH gradually added to 20 cm 3 of 0.1 mol dm -3 H 2 SO 4. H 2 SO 4 is a dibasic acid. It reacts with NaOH in 2 steps. Hence it has 2 end-points. H 2 SO 4 + NaOH NaHSO 4 + H 2 O NaHSO 4 + NaOH Na 2 SO 4 + H 2 O
The 1 st end-point detected after 20 cm 3 of NaOH added can be detected using methyl orange. A solution of NaHSO 4 is left in the titration flask. The 2 nd end-point detected after 40 cm 3 of NaOH added can be detected using phenolphthalein. A solution of NaSO 4 is formed.
E.g 2 : 0.1 mol dm -3 NaOH gradually added to 20 cm 3 of 0.1 mol dm -3 H 3 PO 4. Phosphoric acid is tribasic. It reacts with NaOH in 3 steps. Hence 3 end-points are detected. H 3 PO 4 + NaOH NaH 2 PO 4 + H 2 O NaH 2 PO 4 + NaOH Na 2 HPO 4 + H 2 O Na 2 HPO 4 + NaOH Na 3 PO 4 + H 2 O
Solubility product When an ionic compound, A n B m, is slightly soluble in water, some solid dissolves to form a saturated solution. A n B m (s) nA + (aq) + mB - (aq) When the solution is saturated, the mixture is at equilibrium at a given temperature. K sp = [A + ] n.[B - ] m K sp is called solubility product of the ionic solid. Solubility product only apply to slightly soluble ionic compounds. The K sp expression shows the interaction between ions in the solution. More soluble salts, higher K sp value Less soluble salts, lower K sp value. K sp value are only affected by temperature. Temperature increase, K sp value increase.
Calculating K sp from concentrations. Example 1 : The solubility of calcium sulphate, CaSO 4 at 298K is 0.67 g dm -3. Calculate the solubility product at this temperature.
Example 2 : The solubility of lead (II) chloride, PbCl 2 is mol dm -3 at 298K. Calculate the solubility product at this temperature.
Calculating solubility from K sp Example 1 : the solubility product of calcium carbonate, CaCO 3 = 5.0 x mol 2 dm -6. Calculate the solubility of CaCO 3 in g dm -3
Example 2 :Calculate the solubility of silver chloride, AgCl at 298K if the solubility product is 1.8 x mol 2 dm -6.
Example 3 : Calculate the solubility in g dm -3 of chromium (III) hydroxide, Cr(OH) 3, at 25 C if its solubility product is 1.0 x mol 4 dm -12.
Common ion effect The solubility of an ionic compound of a solution is decreased if the solution already contains one of the ions. Example 1 : A saturated solution of calcium sulphate. CaSO 4 (s) Ca 2+ (aq) + SO 4 2- (aq) K sp = [Ca 2+ ][SO 4 2- ] = 2.4 x mol 2 dm -6 at 298K When a solution that contains the same ion is added to the saturated CaSO 4 solution, for e.g. dilute H 2 SO 4 equilibrium shift to left, CaSO 4 will precipitate out (or solubility decrease). In presence of the additional common ion (SO 4 2- ) the CaSO 4 becomes less soluble. If a solution that contains Ca 2+ ions is used, it will have the same effect. CaSO 4 will precipitate out.
Example 2 : Solubility of AgCl in dilute HCl is less than in pure water because of the common ion, Cl - ion from the HCl. AgCl(s) Ag + (aq) + Cl - (aq) HCl(aq) H + (aq) + Cl - (aq) The presence of Cl - ion from HCl cause the equilibrium to shift left decrease in solubility of AgCl.
Calculation on the common ion effect. E.g 1 : K sp for AgCl is 2.0 x mol 2 dm -6. Calculate the solubility of AgCl in (a) water (b) 0.1 mol dm -3 HCl
E.g 2 : Calculate the solubility of calcium sulphate in mol dm -3 in (a) water (b) 0.50 mol dm -3 dilute H 2 SO 4 K sp = 2.4 x mol 2 dm -6
Uses of K sp 1) Predicting precipitation using K sp K sp can be used to predict whether precipitates will form when solutions are mixed. For a sparingly soluble salt, A x B y : A x B y (s) xA y+ (aq) + yB x- (aq) If [A y+ ] x [B x- ] y = K sp solution is saturated. If [A y+ ] x [B x- ] y < K sp solution is not saturated, no ppt formed. If [A y+ ] x [B x- ] y >K sp ppt forms. [A y+ ] x [B x- ] y is called ionic product.
E.g : Will a precipitate of PbCl 2 be formed if 10 cm 3 of 0.10 mol dm -3 Pb(NO 3 ) 2 is mixed with 10 cm 3 of 0.20 mol dm -3 HCl? K sp (PbCl 2 ) = 1.6 x mol 3 dm -9 at 298K.
Exercise 1. Will a precipitate of Ca(OH) 2 form if 5.0 cm 3 of mol dm -3 NaOH solution is added to 5.0 cm 3 of mol dm -3 CaCl 2 solution? K sp (Ca(OH) 2 ) = 5.5 x mol 3 dm -9 at 298K.
2. Will a precipitate of Ca(OH) 2 form if 5.0 cm 3 of NH 3 solution containing 2.0 x mol dm -3 of OH - ions is added to 5.0 cm 3 of mol dm -3 CaCl 2 solution? K sp (Ca(OH) 2 ) = 5.5 x mol 3 dm -9 at 298K.