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CHEMISTRY 225 SEMESTER 01-2013 HOMEWORK #5 – Answer Key

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CHEMISTRY 225 SEMESTER 01-2013 HOMEWORK #5 – Answer Key CHEMISTRY 225 SEMESTER 01-2013 HOMEWORK #5 – Answer Key 1) A buffer solution is made up by dissolving 8.40 g of sodium dihydrogenphosphate (NaH 2PO 4) and 6.20 g of 3 sodium monohydrogenphosate (Na 2HPO 4) to make 500 cm of solution. Write down the relevant chemical - equilibrium and calculate the pH of the solution, given that pK a(H 2PO4 ) = 7.21. (Only the equilibrium between the dihydrogenphosphate ion and its conjugate base need be considered. Use Na=23, H=1, P=31 and O=16.) Solution - - + The equilibrium to consider is H 2PO 4 (aq) + H 2O(l) l HPO 4 (aq) + H 3O (aq) acid base Cb We apply the buffer equation: pH = pK a + Log 10 but we must calculate C b and C a: Ca -1 Molar mass of NaH 2PO 4 = 23 + 2×1 + 31+ 4×16 =120 g mol 4.8 g ∴ # of mol of NaH 2PO 4 = = .0 0700 mol 120 g mol −1 - ∴ # of mol of H 2PO 4 = 0.0700 mol - .0 0700 mol −3 ∴ Molarity of H 2PO 4 = .0 140 mol dm = 0.140 M = C 5.0 dm 3 a -1 Molar mass of Na 2HPO 4 = 2×23 + 1 + 31+ 4×16 =142 g mol 2.6 g # of mol of Na 2HPO 4 = = 0.043662.. .mol 142 g mol −1 2- ∴ # of mol of HPO 4 = 0.043662…mol 2- .0 043662 ... mol −3 ∴ Molarity of HPO 4 = = 0.087324 mol dm = 0.087324 M = C 5.0 dm 3 b and substituting in the buffer equation gives: .0 087324 pH = 21.7 + Log = 21.7 − .0 204995 ... = 7.005005.. = 7.01 to 2 d.p. 10 0.14 C (Note that we really didn’t need the volume. Since we calculated b , the volume cancelled.) Ca 2) Develop balanced half-equations for - a) the oxidation of iodine (I 2) to iodate (IO 3 ) and Solution - I2 → IO 3 - I2 → 2IO 3 - I2 + 6H 2O → 2IO 3 - + I2 + 6H 2O → 2IO 3 + 12H - + - I2 + 6H 2O → 2IO 3 + 12H + 10e - + - I2(s) + 6H 2O(l) → 2IO 3 (aq) + 12H (aq)+ 10e ……① b) the reduction of nitrate ion to nitrogen monoxide, both in acid solution. Solution - NO 3 → NO - NO 3 → NO + 2H 2O - + NO 3 + 4H → NO + 2H 2O - + - NO 3 + 4H + 3e → NO + 2H 2O - + - NO 3 (aq) + 4H (aq) + 3e → NO(g) + 2H 2O(l) ……... ② Assemble these half-equations into a balanced ionic equation for the oxidation of iodine to iodate by nitrate ion. Indicate the oxidising agent and the reducing agent. Solution Each half-equation is multiplied by a number so as to equalise the number of electrons involved, and then the resulting equations are added. Hence we multiply ① by 3 and ② by 10 to give: - + - 3I 2(s) + 18H 2O(l) → 6IO 3 (aq) + 36H (aq)+ 30e and - + - 10NO 3 (aq) + 40H (aq) + 30e → 10NO(g) + 20H 2O(l) Adding these gives: - + - - + - 3I 2(s) + 18H 2O(l) + 10NO 3 (aq) + 40H (aq) + 30e → 6IO 3 (aq) + 36H (aq)+ 30e + 10NO(g) + 20H 2O(l) and cancelling species that appear on both sides gives us: - + - 3I 2(s) + 10NO 3 (aq) + 4H (aq) → 6IO 3 (aq) + 10NO(g) + 2H 2O(l) - Since the I 2 is oxidised, it is the reducing agent . Since the NO 3 is reduced, it is the oxidising agent . (NB. Always check equations for balance!) 3) Write down the dissolution equilibrium for silver chromate, Ag 2CrO4, and calculate its molar solubility in -12 both water and 0.020 M silver nitrate solution, given that K s(Ag 2CrO4) = 3.0×10 . Solution + 2- The dissolution equilibrium is Ag 2CrO 4(s) l 2Ag (aq) + CrO 4 (aq) 2- In water, since the dissolution 1 mol of Ag 2CrO 4(s) yields 1 mol of CrO 4 in solution, the solubility of 2- Ag 2CrO 4, S(Ag 2CrO 4), or simply S, = [CrO 4 ]. Also since the dissolution of 1 mol of Ag 2CrO 4(s) yields 2 mol of Ag +, [Ag +] = 2S. Hence: = + 2 2− = 2 × = 3 K s [Ag ] [CrO 4 ] 2( S) S 4S K 0.3 ×10 −12 ∴S = 3 s = 3 = 9.0856 ×10 5- = 1.9 ×10 5- M to 2 s. f . 4 4 For the solubility in 0.020 M silver nitrate solution it is best to construct an ICE table. The initial concentration of Ag + is not zero, as it would be in pure water, it is 0.02 M, since it is present in the 2- AgNO 3(aq). However, CrO 4 is only provided by the Ag 2CrO 4, so its initial concentration is zero. + 2- Ag 2CrO 4(s) l 2Ag (aq) + CrO 4 (aq) Initial /M ----- 0.02 0 Change /M ----- +2 x +x Equilibrium /M ----- 0.02+2x x = + 2 2− = − 2 Now K s [Ag ] [CrO 4 ] 02.0( 2x) (x) We assume that 2x<<0.02, so that 0.02+2 x ≈ 0.02. Thus: = 2 = × −4 K s 02.0 x 4 10 x −12 K 3×10 − ∴x = s = = 5.7 ×10 9 4 ×10 −4 4 ×10 −4 2- Once again, [CrO 4 ] = S(Ag 2CrO 4) = x and so: -9 S(Ag 2CrO 4) = 7.5×10 M To check the approximation, we compare 2x with 0.02: −9 −5 2x 2× 5.7 ×10 − 75.3 ×10 − = = 5.7 ×10 7 = = 5.7 ×10 5% 0.02 0.02 100 That is, 2x is 7.5×10 -5% of 0.02, so it is clearly much less than 5% of 0.02, and the approximation is acceptable. Suggest a reagent which could be used to increase the solubility of silver chromate. Solution Anything which removes one of the product ions from solution will do. Ammonia forms a complex with + + + 2- - the Ag , [Ag(NH 3)2] , and hydrogen ion, H , will react with the CrO 4 , which is basic, to form HCrO 4 , and then H 2CrO 4. Hence silver chromate is soluble in ammonia solution or acid solution. That is, either ammonia or, for example, dilute nitric acid, will increase the solubility of silver chromate. (Dilute hydrochloric acid or dilute sulfuric acid would not work because silver chloride or silver sulfate would be precipitated.) .
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