Chemistry Labs

Problem 1

(1.1) Show that 0.1 mol of TlX2S\ce{Tl2S} dissolves in one litre of a 1 M solution of any strong monoprotic non-coordinating acid. (1.2) Show that 0.1 mol of CuS\ce{CuS} dissolves in 1 M HNOX3\ce{HNO3} but not in 1 M HCl\ce{HCl}. Data: E0(S/SX2−)=−0.48E^0(\ce{S/S^{2-}}) = -0.48 V; E0(NOX3X−/NO)=0.96E^0(\ce{NO3^-/NO}) = 0.96 V; pKa(HX2S)=7K_a(\ce{H2S}) = 7; pKa(HSX−)=13K_a(\ce{HS^-}) = 13; Ksp(TlX2S)=1×10−20K_{sp}(\ce{Tl2S}) = 1\times10^{-20}; Ksp(CuS)=1×10−35K_{sp}(\ce{CuS}) = 1\times10^{-35}; solubility of NO\ce{NO} in water (298 K): 2.53×10−22.53\times10^{-2} mol dm−3^{-3}. Assume that CuX2+\ce{Cu^{2+}} does not form stable complexes with ClX−\ce{Cl^-}.
Step 2 of 4: Tl₂S dissolves
[TlX+]2[SX2−]=(0.2)2×10−21=4×10−23<Ksp(TlX2S)=10−20[\ce{Tl+}]^2[\ce{S^{2-}}] = (0.2)^2 \times 10^{-21} = 4\times10^{-23} < K_{sp}(\ce{Tl2S}) = 10^{-20}
Analysis

If 0.1 mol TlX2S\ce{Tl2S} dissolves per litre, [TlX+]=0.2[\ce{Tl+}] = 0.2 M and total sulphide c=0.1c = 0.1 M. The ion product 4×10−234\times10^{-23} is below Ksp=10−20K_{sp} = 10^{-20}, so dissolution in 1 M strong acid is spontaneous.