sign in sign up
<<
Home , Silver chromate

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 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 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 chromate, Ag 2CrO4, and calculate its molar in -12 both water and 0.020 M 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 from solution will do. 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 , will increase the solubility of silver chromate. (Dilute hydrochloric acid or dilute sulfuric acid would not work because silver or silver sulfate would be precipitated.)