pH of 0.100 M ammonium sulfide. What is the pH of a 0.100M aqueous solution of (

Question

pH of 0.100 M ammonium sulfide. What is the pH of a 0.100M aqueous solution of (NH4)2S? An exact analytical solution to this question can be algebraically very complex, however we can use some chemical intuition to greatly simplify it. Here is a method to approach this question. We know that the solution equilibrium will lie somewhere along the path: H2S + 2NH3 HS-+ NH4+ + NH3 S2-+ 2NH4 + But, where?? Consider the following equilibria: H2s(aq) + H20(I) HS-(aq) + H3O+(aq) K1 = Ka,H2S = 1.0×10-7 NH3(aq) + H2O(l) H30+(aq) + OH-(aq) NH4+(aq) + OH-(aq) K2 = KOMMy = 1.8×10-5 2H20(l) K3 = 1/Kw = 1.0× 1014 Combine the above equations to get: H2S(aq) + NH3(aq) HS-(aq) + NH4+(aq) Keq-K1K2K3-1.8x102 This relatively large equilibrium constant suggests that, when near stoichiometric amounts of the reagents are present in an aqueous solution, the equilibrium will favour the products. Now, since H2S is diprotic we need to briefly consider the second dissociation step: HS-(aq) + H2O(l) S2-(aq) + H30"(aq) K4 = Ka,HS-1.0x10-19 NH3(aq) + H2O(1) NH4+(aq) + OH-(aq) K2 Kb,NH,-1.8x 10-5 H30+(aq) + OH-(aq)-2H2O(l) K3 # 1/Kw = 1.0x1014 Combine the above equations to get: HS-(aq) + NH3(aq) S2-(aq) + NH4+(aq) Keq2-K4K2K3 = 1.8x10-10

Explanation / Answer

[OH-] = K2[NH3]/ [NH4+] , [NH3] = 0.1+0.1/180 =0.100556

[NH4+] = 0.1-1/180= 0.0944

[OH-] = 1.8*10-5*0.1000556/0.0944=1.92*10-5

poH= -log (1.92/10^5)= 4.72

pH= 14-4.72= 9.28

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