Problem 1

(1.1) Show that 0.1 mol of

\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

\ce{CuS}

dissolves in 1 M

\ce{HNO3}

but not in 1 M

\ce{HCl}

. Data:

E^0(\ce{S/S^{2-}}) = -0.48

E^0(\ce{NO3^-/NO}) = 0.96

V; p

K_a(\ce{H2S}) = 7

; p

K_a(\ce{HS^-}) = 13

;

K_{sp}(\ce{Tl2S}) = 1\times10^{-20}

;

K_{sp}(\ce{CuS}) = 1\times10^{-35}

; solubility of

\ce{NO}

in water (298 K):

2.53\times10^{-2}

mol dm

^{-3}

. Assume that

\ce{Cu^{2+}}

does not form stable complexes with

\ce{Cl^-}

Step 4 of 4: Nitric acid oxidises sulphide

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Intuition

Protonation alone lowers [S²⁻] by 20 orders; oxidation removes S²⁻ essentially completely, which is what a salt with Ksp = 10⁻³⁵ needs.

2\ce{NO3^-} + 8\ce{H+} + 3\ce{S^{2-}} -> 3\ce{S} + 2\ce{NO} + 4\ce{H2O};\quad \Delta E^0 = 0.96 + 0.48 = 1.44\ \mathrm{V};\quad \log K = \frac{6 \times 1.44}{0.059} \Rightarrow K \approx 10^{144}

Analysis

In $\ce{HNO3}$ the nitrate oxidises $\ce{S^{2-}}$ to S with $\Delta E^0 = 1.44$ V for a six-electron reaction, giving $K \approx 10^{144}$ . With $[\ce{NO3^-}] = [\ce{H+}] = 1$ , $[\ce{S^{2-}}]^3 = [\ce{NO}]^2/K$ ; taking the saturated $[\ce{NO}] = 0.0253$ M gives $[\ce{S^{2-}}] \approx 10^{-51}$ M, so $[\ce{Cu^{2+}}][\ce{S^{2-}}] \approx 10^{-52} \ll K_{sp}$ and CuS dissolves.