CHEMISTRY SK015 2023/2024 SUGGESTED ANSWER TUTORIAL 7.0: IONIC EQUILIBRA 1 7.1 Acids and Bases 1. Define acid and base according to the Arrhenius, Bronsted-Lowry and Lewis theories. Write a suitable example by using a different equation for each theory. Arhenius Theory Acid is a substance that dissociates in aqueous solution/water to produce hydrogen ion or hydronium ion in aqueous solution. Example : HCl (s) + H2O (l) → H3O + (aq) + Cl- (aq) Base is a substance that dissociates in aqueous solution/water to produce hydroxide ion. Example : NH3 (s) + H2O (l) → NH4 + (aq) + OH- (aq) Bronsted-Lowry Theory Acid is a substance that can donate a proton (H+ ) to another substance. Example : HNO3 (aq) + H2O (l) → H3O + (aq) + NO3 - (aq) Base is a substance that can accept a proton (H+ ) from another substance. Example : CO3 2- (aq) + H2O (l) → HCO3 - (aq) + OH- (aq) Lewis Theory Acid is a substance that can accept a pair of electrons to form coordinate covalent bond. Base is a substance that can doanate a pair of electrons to form coordinate covalent bond. Example : NH3 + BF3 → NH3-BF3 base acid 2. For each of the following, identify the conjugate acid-base pairs : (a) NH3 + H2PO4 - NH4 + + HPO4 2- (b) HClO + CH3NH2 CH3NH3 + + ClO-
CHEMISTRY SK015 2023/2024 SUGGESTED ANSWER TUTORIAL 7.0: IONIC EQUILIBRA 2 Conjugate pairs Acid Conjugate Base Base Conjugate Acid a) H2PO4 - HPO4 2- NH3 NH4 + b) HClO ClO- CH3NH2 CH3NH3 + 3. (a) Define pH. pH is the negative logarithm of the concentration of hydrogen ion. pH = - log [H+ ] (b) Calculate the pH of 0.05 M of hydrochloric acid solution. pH = - log (0.05) = 1.30 (c) The concentration of hydroxide ion in a blood sample is 2.5 x 10-7 M. What is the pH of the blood? pOH = −log [OH− ] pOH = −log (2.5 x 10-7 ) = 6.60 pH = 14 − 6.60 pH = 7.40 4. Calculate the mass of NaOH needed to prepare 500.0 mL of solution with a pH of 10.00. pOH = 14 − 10 = 4 pOH = −log [OH− ] [OH- ] = 1.010−4 M The concentration of OH− = number of moles NaOH /0.500 L mol NaOH needed = 1 x 10−4 0.5 = 5.00 10−5 mol mass NaOH = 5.00 10−5 mol 40 (g/mol) = 0.002 g 5. (a) Write an expression for the dissociation constant, Ka of propanoic acid, CH3CH2COOH.
CHEMISTRY SK015 2023/2024 SUGGESTED ANSWER TUTORIAL 7.0: IONIC EQUILIBRA 3 (b) Calculate the pH of a 0.35 mol L−1 solution of propanoic acid at 25oC. [Ka = 1.3510−5 mol L-1 ] CH3CH2COOH(aq) + H2O (l) CH3CH2COO- (aq) + H3O + (aq) [ ]initial 0.35 0 0 [ ] −x +x +x [ ]eq 0.35 − x x x = x x 0.35 − 2 = 1.35 x 10−5 Ka<<1, 0.35 − x 0.35, 0.35 2 x = 1.3510−5 x = 2.17310−3 M = [H+ ] pH = −log [H+ ] = − log (2.17310−3 ) = 2.66 6. The pH of a 0.036 M solution of nitrous acid, HNO2, is 2.40. What is the Ka of the acid? pH = −log [H3O + ] = 2.40 [H+ ] = 3.9810−3 M HNO2(aq) + H2O(l) H3O + (aq) + NO2 - (aq) [ ]initial 0.036 0 0 [ ] −3.9810−3 +3.9810−3 +3.9810−3 [ ]eq 0.10 – 3.9810−3 3.9810−3 3.9810−3 Ka = [NO2 - ][H3O + ] [HNO2] = (3.98 x 10-3 ) 2 (0.036 – 3.98 x 10-3 ) = 4.95 10−4 Ka = Ka =
CHEMISTRY SK015 2023/2024 SUGGESTED ANSWER TUTORIAL 7.0: IONIC EQUILIBRA 4 7. Calculate the percentage dissociation and the degree of dissociation,α of 0.10 M hydrocyanic acid, HCN solution. [Ka HCN = 5.0 10−10 mol L−1 ] . HCN(aq) + H2O(l) H3O + (aq) + CN− (aq) [ ]initial 0.10 0 0 [ ] −x +x +x [ ]eq 0.10 − x x x Ka = [CN- ][H3O + ] = 5.0 10−10 [HCN] = x 2 = 5.0 10 −10 0.10- x Assuming that 0.10 – x 0.10, we get x 2 = 5.0 10−10 0.10 x = 7.07 x 10-6 M Thus, % dissociation = 7.07 x 10-6 M 100% 0.10 M = 7.1 x 10-3 % Degree of dissociation, α = 7.07 x 10-6 M 100% 0.10 M = 0. 71 8. The percentage ionisation of 0.010 M NH3 solution was 4.2 % ionisation. Calculate Kb. NH3 (aq) + H2O (l) NH4 + (aq) + OH− (aq) [ ]initial 0.01 0 0 [ ] − 4.210−4 + 4.210−4 + 4.210−4 [ ]eq 9.5810−3 4.210−4 4.210−4 Kb = [NH4 + ][ OH- ] [NH3]
CHEMISTRY SK015 2023/2024 SUGGESTED ANSWER TUTORIAL 7.0: IONIC EQUILIBRA 5 Kb = (4.210-4 )(4.210-4 ) = 1.810-5 9.58 10-3 9. What is the pH of a 0.10 M solution of phenylamine, C6H5NH2? [Kb C6H5NH2 = 4.0 10−10 mol L−1 ]. C6H5NH2 (aq) + H2O(l) C6H5NH3 + (aq) + OH− (aq) [ ]initial 0.10 0 0 ∆[ ] − x +x +x [ ]eq 0.10 − x x x Kb = [C6H5NH3 + ][ OH− ] = 4.010−10 [C6H5NH2] x 2 = 4.010−10 0.10- x Assuming 0.10 – x 0.10, because of Kb << 1, x 2 = 4.0 10−10 0.10 x = 6.32 x 10-6 M = [OH- ] pH = 14 − (-log [OH− ]) = 14 − (-log 6.32 10−6 ) = 8.3 10. The table shows the base ionisation constant, Kb, for several selected compounds. (a) Arrange the compounds in order of increasing strength of base. C6H5NH2 < NH2OH < N2H4 < NH3 (increasing strength of base) Compound Kb C6H5NH2 3.810−10 N2H4 1.710−6 NH3 1.810−5 NH2OH 1.110−8
CHEMISTRY SK015 2023/2024 SUGGESTED ANSWER TUTORIAL 7.0: IONIC EQUILIBRA 6 (b) Give the structure of conjugate acid for each compound and arrange them in order of increasing strength of acid. NH4 + < N2H5 + < NH3OH+ < C6H5NH3 + (increasing strength of acid) 11. Write the salt hydrolysis equation for the following salts and classify them as acidic, basic or neutral. (a) NaCN Answer: Basic salt NaCN → Na+ + CNHydrolysis equation for cation : CN- (aq) + H2O(l) HCN(aq) + OH− (aq) (b) N2H5Cl Answer: Acidic salt N2H5Cl → N2H5 + + ClHydrolysis equation for cation : N2H5 + (aq) + H2O(l) N2H4(aq) + H3O + (aq) 12. (a) Calculate the pH of a solution containing 0.20 M CH3COOH and 0.30M CH3COONa. [Ka CH3COOH = 1.8 x 10-5 ] Answer: By using Henderson-Hasselbalch equation, the pH of the solution pH = pKa + log [CH3COONa] [CH3COOH] = −log 1.810−5 + log (0.30/0.20) = 4.9 (b) Calculate the pH of the 0.20 M CH3COOH solution if there is no salt present. [Ka CH3COOH = 1.810−5 ] Answer: Without the salt, it is just a weak acid.
CHEMISTRY SK015 2023/2024 SUGGESTED ANSWER TUTORIAL 7.0: IONIC EQUILIBRA 7 CH3COOH(aq) + H2O (l) CH3COO− (aq) + H3O + (aq) [ ]initial 0.20 0 0 [ ] −x +x +x [ ]eq 0.20 −x x x Ka = [H3O + ][CH3COO- ] = 1.810−5 [ CH3COOH] x 2 = 1.810−5 0.20-x Applying the approximation 0.20 − x ≈ 0.20 x 2 = 1.810−5 0.20 x = 1.89710−3 pH = −log[H+ ] = −log (1.89710−3 ) = 2.7 (c) Explain the change in pH the solution in (a) when i. a small amount of strong acid is added. Answer: When a small amount of strong acid is added to the mixture of solution, the H + from the acid reacts with the CH3COOto form CH3COOH. CH3COO− (aq) + H + (aq) CH3COOH(aq) As a result, the H+ ion is consumed and hence the pH of the solution is not much affected. ii. a small amount of strong base is added. Answer: When a small amount of strong base is added to the mixture of solution, the OHfrom the base reacts with the CH3COOHto form CH3COO- . CH3COOH(aq) + OH− (aq) CH3COO− (aq) + H2O(aq) As a result, the hydroxide ions are consumed and hence the pH of the solution is not much affected.
CHEMISTRY SK015 2023/2024 SUGGESTED ANSWER TUTORIAL 7.0: IONIC EQUILIBRA 8 13. A student is asked to prepare a buffer solution at pH 4.6 using 50.0 mL of 0.5 M benzoic acid, C6H5COOH and sodium benzoate, C6H5COONa. Calculate the mass of sodium benzoate required to prepare the buffer solution. [Ka C6H5COOH = 6.5 x 10-5 ] Answer: pH = pKa + log [C6H5COO- ] [C6H5COOH] pKa = - log 6.5 x 10-5 = 4.18 log [C6H5COO- ] = 4.6 – 4.18 = 0.42 [C6H5COOH] [C6H5COO- ] = 2.63 [C6H5COOH] [C6H5COO- ] = 2.63 x 0.5 = 1.32 M nsalt = 1.32 x 50/100 = 6.6 x 10-2 mass salt = 6.6 x 10-2 x 144 = 9.50 g 14. A buffer solution was prepared by dissolving 0.15 mol of sodium ethanoate in 1 dm3 of 0.10 M ethanoic acid. (a) Calculate the pH of the buffer solution above. pH = 4.74 – log _0.1_ 0.15 = 4.92 (b) Calculate the change in pH when 1.0 cm3 of 1.0 M hydrochloric acid is added to 100 cm3 of the buffer solution. No moles of H+ added = 0.001 dm3 x 1 M = 0.001 mol No moles of CH3COOH in 100 cm3 = 0.01 mol No moles of CH3COOin 100 cm3 = 0.015 mol CH3COO- + H+ CH3COOH
CHEMISTRY SK015 2023/2024 SUGGESTED ANSWER TUTORIAL 7.0: IONIC EQUILIBRA 9 Using ICF table, CH3COO- H + CH3COOH Initial/mol 0.015 0.001 0.01 Change/mol -0.001 -0.001- +0.001 Final/ mol 0.014 0 0.011 [CH3COOH] final = 0.011 mol x 0.101 dm3 = 0.1089 M [CH3COO- ] final = 0.014 mol x 0.101 dm3 = 0.1386 M So, final pH will be , = 4.74 – log 0.1089 0.1386 = 4.84 Hence, change in pH =Initial pH – Final pH = 4.92 – 4.84 = 0.08 15. Calculate pH of the solution formed when: (a) 150 mL of 0.200 M HNO3 is mixed with 75.0 mL of 0.200 M NaOH. Answer: HNO3(aq) + NaOH(aq) ⎯→ H2O(aq) + NaNO3(s) ninitial 1000 150.00x0.2 = 0.03 mol 1000 75.00x0.2 = 0.015 mol - n −0.015 mol −0.015 mol +0.015 mol nfinal 0.015 mol 0 0.015 mol The total volume is 150 + 75 = 225 mL [H+ ] = L mol 0.225 0.015 = 0.0667 M pH = − log [H+ ]
CHEMISTRY SK015 2023/2024 SUGGESTED ANSWER TUTORIAL 7.0: IONIC EQUILIBRA 10 = − log 0.0667 = 1.18 (b) 25.0 mL of 0.450 M H2SO4 is mixed with 25.0 mL of 0.900 M NaOH. Answer: 2NaOH(aq) + H2SO4(aq) ⎯→ 2H2O(aq) + Na2SO4(aq) ninitial 25.0 x 0.9 1000 = 0.0225 mol 25.0 x 0.45 1000 = 0.01125 mol - n −0.0225mol -0.01125 mol +0.01125 mol nfinal 0 0 0.01125 mol Since there is no excess acid and base and the solution is left with a neutral salt , Na2SO4 and water. Therefore, pH of the solution = 7 at 25C. 7.2 Acid –base Titration 16. a) A titration experiment at 25oC involving a total of 30 mL of 0.1 M NaOH and 25mL of 0.1 M of HNO3. NaOH was added dropwise along the experiment. Show the variation of pH of the solution i) before the addition of NaOH, ii) at half equivalence point, iii) at equivalence point and iv) at final volume of the titration. b) Sketch a suitable graph for the titration process. Choose the most suitable indicator based on the following table and give reason for your choice. Indicator pH range Phenoplhtalein 8.3 - 10.0
CHEMISTRY SK015 2023/2024 SUGGESTED ANSWER TUTORIAL 7.0: IONIC EQUILIBRA 11 Cresol red 7.0 – 8.8 Bromothymol blue 6.0 – 7.6 [given Kw at 25 oC = 1 x 10-14 a) Answer: i) HNO3 + NaOH → NaNO3 + H2O Before the addition of NaOH, pH = - log [H+ ] @ = - log [0.1] = 1.0 ii) At half equivalence point, Volume of NaOH = 12.5 mL Mol of NaOH = 12.5 x 0.1 M = 1.25 mmol HCl + NaOH → NaCl + H2O Initial mol 2.5 1.25 Δ -1.25 -1.25 + 1.25 Final 1.25 0 1.25 pH = - log = - log (0.033 M) = 1.5 iii) At equivalence point, Mol NaOH = mol HNO3 1.25 37.5
CHEMISTRY SK015 2023/2024 SUGGESTED ANSWER TUTORIAL 7.0: IONIC EQUILIBRA 12 HNO3 + NaOH → NaNO3 + H2O Initial mmol 25 x 0.1 25 x 0.1 0 Δ -2.5 -2.5 + 2.5 Final 0 0 2.5 At equivalence point, H2O H+ + OHKw = [H+] [OH- ] = 1 x 10-14 @ [H+ ] = 1 x 10-7 pH = - log [H+ ] = 7.0 @ 7 iv) 30 mL NaOH added (final), HNO3 + NaOH → NaNO3 + H2O Initial mmol 25 x 0.1 = 2.5 30 x 0.1 = 3.0 0 Δ -2.5 -2.5 + 2.5 Final 0 0.5 2.5 pOH = - log [OH- ] = - log = - log (9.09 x 10-3 ) = 2.04 pH = 14 – 2.04 = 12.0 0.5 30 + 25
CHEMISTRY SK015 2023/2024 SUGGESTED ANSWER TUTORIAL 7.0: IONIC EQUILIBRA 13 The most suitable indicator is Bromothymol blue : - The range covers the end point of this titration - The starting range just before the end point - Drastic or abrupt colour change 7.3 Solubility equilibria 17. (a) State the differences between solubility and solubility product. Solubility is the amount of substance that is dissolved in a known volume of a solution. Solubility product is the product of the solubilities of the ions in a saturated solution. (b) Calculate the solubility in g L-1 for calcium fluoride, CaF2. [Ksp CaF2 = 3.98 x 10- 11] CaF2 (s) Ca2+ (aq) + 2F- (aq) x 2x x(2x)2 = 3.98 x 10 -11 4x3 = 3.98 x 10 -11 x = 2.15 x 10 -4 Ksp CaF2 = 3.98 x 10 -11 = [Ca2+] [F- ] 2 ] Molar solubility = 2.15 x 10 -4 mol L-1 Solubility = (2.15 x 10 -4 ) x 78.1 = 0.01679 g L-1 half-equivalence point equivalence point 1.0 7.0 12.0 5.0 10.0 15.0 20.0 25.0 30.0 pH Volume of NaOH (mL) 1 12.5 halfequivalenc final final volume
CHEMISTRY SK015 2023/2024 SUGGESTED ANSWER TUTORIAL 7.0: IONIC EQUILIBRA 14 (c) The solubility of calcium fluoride, CaF2 in water is 0.01679 gL-1 . Calculate the value of Ksp. Molar solubility = 0.01679 gL -1 78.1 = 2.15 x 10 -4 mol L-1 Ksp CaF2 = [Ca2+] [F- ] 2 ] = 4 (2.15 x 10 -4 ) 3 = 3.98 x 10 -11 18. Will precipitate form if 200 mL of 0.004 M BaCl2 is added to 600 mL of 0.008 M K2SO4? [Ksp BaSO4= 1.110−10] Answer: The precipitate formed is BaSO4. BaSO4 (s) Ba2+(aq) + SO4 2− (aq) The number of moles of Ba2+ present in the 200 mL of BaCl2 solution is 200 mL x 0.0040 M = 8.010-4 mol 1000 mL The total volume after combining the two solutions is 800 mL. The concentration of Ba2+ in the 800 mL volume is [Ba2+] = 4 8.0 10 0.80 mol L − = 1.010−3 M The number of moles of SO4 2- in the 600 mL K2SO4 solution is 600 mL 0.008 M = 4.810−3 mol SO4 2− 1000 mL The concentration of SO4 2− in the 800 mL of the combined solution is [SO4 2- ] = 3 4.8 x10 0.80 mol L − = 6.010−3 M Now comparing Q and Ksp. Q = [Ba2+][SO4 2− ]
CHEMISTRY SK015 2023/2024 SUGGESTED ANSWER TUTORIAL 7.0: IONIC EQUILIBRA 15 = (1.010−3 )( 6.010−3 ) = 6.010−6 > 1.110−10 Therefore, Q > Ksp The value of Q indicates that the concentrations of the ions are too large, thus some of the BaSO4 will precipitate out of solution until [Ba2+][ SO4 2− ] = 1.110−10 19. Calculate the solubility of silver chromate, Ag2CrO4 at 25C in (a) liquid water Answer: Ag2CrO4(s) 2 Ag+ (aq) + CrO4 2− (aq) [ ]initial 0 0 ∆[ ] +2s +s [ ]eq 2s s Ksp = [Ag+]2 [CrO4 2- ] 9.010-12 = (2s)2 (s) 4s3 = 9.010−12 s = 1.310−4 M (b) solution of 0.005 M K2CrO4 solution. [Ksp Ag2CrO4 = 9.010−12] Explain the difference in the values of solubility. Answer: Ag2CrO4(s) 2 Ag+ (aq) + CrO4 2− (aq) [ ]initial 0 0.005 ∆[ ] +2y +y [ ]eq 2y y + 0.005 Ksp = [Ag+]2 [CrO4 2- ] 9.0 x 10-12 = (2y)2 (y + 0.005); (y + 0.005)≈ 0.005 4y2 (0.005) = 9.010−12 y = 2.1210−5 M
CHEMISTRY SK015 2023/2024 SUGGESTED ANSWER TUTORIAL 7.0: IONIC EQUILIBRA 16 The solubility of Ag2CrO4 in KCrO4 solution is lower than in water. The presence of the common ion CrO4 - increses the concentration of the CrO4 - ion and thus shifts the equilibrium to the left Hence increase the formation of Ag2CrO4 and decrease/reduce the solubility. 20. (a) The solubility of Ag2SO4 in water at temperature 25C is 0.506 g for 100.00 mL solution. Calculate the solubility product at temperature 25C Solubility of Ag2SO4 = Ag2SO4 2Ag+ + SO4 2- [ ](M) 2(0.01621) 0.01621 2 2 2 5 3 Ksp [SO4 ][Ag ] (0.01621)(0.01621 2) 1.70 10 M − + − = = = (b) If an aqueous solution of Na2SO4 was added progressively into an aqueous solution of 0.01 M Ag2SO4, determine the minimum concentration of Na2SO4 needed just to begin the precipitation of Ag2SO4. Common-ion effect Ag2SO4 2Ag+ + SO4 2- 2(0.01) 0.01 Na2SO4 → 2Na+ + SO4 2- x [ SO4 2- ] = x + 0.01 [ Ag+ ] = 2(0.01) To begin a precipitation, Q ≥ Ksp [ Ag+ ] 2 [ SO4 2- ] ≥ Ksp [ 2(0.01) ] 2 (x + 0.01) ≥ 1.70 X 10-5 x ≥ 0.0325M Therefore, the minimum concentration of Na2SO4 needed just to begin the precipitation of Ag2SO4 is 0.0325M M L gmol g 0.01621 0.1 2(108) 32 64 0.506 1 = + + −