For Fe2+(aq) + S2-(aq) FeS(s), the equilibrium constant at 298 K is Kc = 1.6 * 10¹⁷. Equal volumes of 0.06 M Fe2+(aq) and 0.2 M S2-(aq) are mixed. The equilibrium concentration of Fe2+(aq) equals Y * 10⁻¹⁷ M. Find Y (to two decimal places).

Question

For Fe2+(aq) + S2-(aq) <-> FeS(s), the equilibrium constant at 298 K is Kc = 1.6 * 10¹⁷. Equal volumes of 0.06 M Fe2+(aq) and 0.2 M S2-(aq) are mixed. The equilibrium concentration of Fe2+(aq) equals Y * 10⁻¹⁷ M. Find Y (to two decimal places).

Accepted Answer

8.93. After mixing, [Fe2+]0 = 0.03 M and [S2-]0 = 0.10 M; nearly all Fe2+ reacts leaving excess [S2-] = 0.07 M, so [Fe2+] = 1/(Kc*[S2-]) = 1/(1.6*10¹⁷ * 0.07) = 8.93*10⁻¹⁷ M, giving Y = 8.93.

For Fe2+(aq) + S2-(aq) <-> FeS(s), the equilibrium constant at 298 K is Kc = 1.6 * 10¹⁷. Equal volumes of 0.06 M Fe2+(aq) and 0.2 M S2-(aq) are mixed. The equilibrium concentration of Fe2+(aq) equals Y * 10⁻¹⁷ M. Find Y (to two decimal places).

Solution

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