Question

Chloride ions are analytically determined by precipitating them with silver nitrate. The precipitate is undergoing decomposition in presence of light and forms elemental silver and chlorine. In aqueous solution the latter disproportionates to chlorate(V) and chloride. With excess of silver ions, the chloride ions formed are precipitated whereas chlorate(V) ions are not. 

(a) Write the balanced equations of the reactions mentioned above. 

(b) The gravimetric determination yielded a precipitate of which 12% by mass was decomposed by light. Determine the size and direction of the error caused by this decomposition. 

(c) Consider a solution containing two weak acids HA and HL,0.020 molar and 0.010 molar solutions, respectively. The acid constants are 1 × 10^-4 for HA and 1 ×10^-7 for HL. Calculate the pH of the solution. 

(d) M forms a complex ML with the acid H₂L with the formation constant K₁.The solution contains another metal ion N that forms a complex NHL with the acid H₂L.Determine the conditional equilibrium constant, K'₁ for the complex ML in terms of [H⁺ ] and K values. 

K_t = [ML]/[M][L]

K_t = [ML]/[M][L]

[M'] = total concentration of M not bound in Ml 

[L'] = the sum of the concentrations of all species containing L except ML 

In addition to K₁, the acid constants K_a1 and K_a2 of H₂L as well as the formation constant 

K_NHL of NHL are known. 

K_NHL =[NHL]/[N][L][H⁺]

You may assume that the equilibrium concentration [H⁺] and [N] are known, too. 

Answer 1

(b) From 100 g AgCl 12 g decompose and 88 g remain. 12 g equals 0.0837 mol and therefore, 0.04185 mol Cl₂ are liberated. Out of that (12 ×107.9)/143.3 = 9.03 g Ag remain in the precipitate. 5/6 × 0.837 mol AgCl are newly formed (= 10.0 g), so that the total mass of precipitate (A) yields: A = 88 g + 9.03 g + 10.0 g = 107.03 g; relative error = 7.03 %

(c) [H⁺] = [A^- ] + [L^- ] + [OH^- ]  

[HA]+ [A^- ] = 0.02 mol dm^-3 pK(HA) = pH + p[A^-] -p[HA] = 4 

[HL] + [L^- ] = 0.01 mol dm^-3 pK(HL) = pH + p[L^-] - p[HL] = 7

For problems like these, where no formal algebraic solution is found, only simplifications lead to a good approximation of the desired result, e.g. 

1. [H+ ] = [A- ] (since HA is a much stronger acid than HL then [A^- ] » [L^- ] + [OH^- ]) 

H⁺]²⁺ K(HA)[H⁺ ] – K_(HA)0.02 = 0

[H+ ] = 1.365 × 10^-3 mol dm^-3

 pH = 2.865

2. Linear combination of the equations

The equation above can only be solved by numerical approximation methods. The result is pH = 2.865.We see that it is not necessary to consider all equations. Simplifications can be made here without loss of accuracy.Obviously it is quite difficult to see the effects of a simplification - but being aware of the fact that already the so-called exact solution is not really an exact one (e.g. activities are not being considered), simple assumption often lead to a very accurate result.