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ChemMedAdv 2020 • Paper 2

Minimum Hydrogen Ion Concentration to Prevent Precipitation of Zinc Sulfide

An acidic solution containing 0.05 M Zn2+0.05\text{ M } \text{Zn}^{2+} is saturated with 0.1 M H2S0.1\text{ M } \text{H}_2\text{S}. Compute the minimum molar concentration (M) of H+\text{H}^+ required to suppress the precipitation of ZnS\text{ZnS}. Given parameters: Ksp(ZnS)=1.25×10−22K_{\text{sp}}(\text{ZnS}) = 1.25 \times 10^{-22} and overall dissociation constant of H2S\text{H}_2\text{S}, KNET=K1K2=1×10−21K_{\text{NET}} = K_1 K_2 = 1 \times 10^{-21}.

Official Numerical Answer0.2

Topics & Concepts

EquilibriumSolubility ProductAcid-Base EquilibriumChemical Equilibrium

Step-by-Step Solution

To find the minimum molar concentration of H+\text{H}^+ required to prevent the precipitation of ZnS\text{ZnS}, we need to ensure that the ionic product of ZnS\text{ZnS} does not exceed its solubility product constant (KspK_{\text{sp}}).

Step 1: Determine the maximum allowable concentration of S2−\text{S}^{2-} ions

The dissociation equilibrium for ZnS\text{ZnS} is given by: ZnS(s)⇌Zn2+(aq)+S2−(aq)\text{ZnS}(s) \rightleftharpoons \text{Zn}^{2+}(aq) + \text{S}^{2-}(aq)

The solubility product expression is: Ksp(ZnS)=[Zn2+][S2−]K_{\text{sp}}(\text{ZnS}) = [\text{Zn}^{2+}][\text{S}^{2-}]

To prevent precipitation, the condition to be satisfied is: [Zn2+][S2−]≤Ksp(ZnS)[\text{Zn}^{2+}][\text{S}^{2-}] \le K_{\text{sp}}(\text{ZnS})

Substitute the given values: [Zn2+]=0.05 M[\text{Zn}^{2+}] = 0.05\text{ M} Ksp(ZnS)=1.25×10−22K_{\text{sp}}(\text{ZnS}) = 1.25 \times 10^{-22}

Thus, the maximum concentration of S2−\text{S}^{2-} permitted in the solution is: [S2−]max=Ksp(ZnS)[Zn2+]=1.25×10−220.05=2.5×10−21 M[\text{S}^{2-}]_{\text{max}} = \frac{K_{\text{sp}}(\text{ZnS})}{[\text{Zn}^{2+}]} = \frac{1.25 \times 10^{-22}}{0.05} = 2.5 \times 10^{-21}\text{ M}

Step 2: Relate [S2−][\text{S}^{2-}] to [H+][\text{H}^+] using the overall dissociation of H2S\text{H}_2\text{S}

The overall dissociation equilibrium of H2S\text{H}_2\text{S} in aqueous solution is: H2S(aq)⇌2H+(aq)+S2−(aq)\text{H}_2\text{S}(aq) \rightleftharpoons 2\text{H}^+(aq) + \text{S}^{2-}(aq)

The overall equilibrium constant expression KNETK_{\text{NET}} is: KNET=K1K2=[H+]2[S2−][H2S]K_{\text{NET}} = K_1 K_2 = \frac{[\text{H}^+]^2 [\text{S}^{2-}]}{[\text{H}_2\text{S}]}

Rearranging to express [H+]2[\text{H}^+]^2: [H+]2=KNET×[H2S][S2−][\text{H}^+]^2 = \frac{K_{\text{NET}} \times [\text{H}_2\text{S}]}{[\text{S}^{2-}]}

Step 3: Calculate the minimum concentration of H+\text{H}^+

Given:

  • KNET=1×10−21K_{\text{NET}} = 1 \times 10^{-21}
  • [H2S]=0.1 M[\text{H}_2\text{S}] = 0.1\text{ M}
  • [S2−]max=2.5×10−21 M[\text{S}^{2-}]_{\text{max}} = 2.5 \times 10^{-21}\text{ M}

Substitute these values into the equation: [H+]2=(1×10−21)×0.12.5×10−21[\text{H}^+]^2 = \frac{(1 \times 10^{-21}) \times 0.1}{2.5 \times 10^{-21}}

[H+]2=1×10−222.5×10−21=0.12.5=0.04 M2[\text{H}^+]^2 = \frac{1 \times 10^{-22}}{2.5 \times 10^{-21}} = \frac{0.1}{2.5} = 0.04\text{ M}^2

Taking the square root on both sides: [H+]=0.04=0.2 M[\text{H}^+] = \sqrt{0.04} = 0.2\text{ M}

Final Answer

The minimum molar concentration of H+\text{H}^+ required to prevent the precipitation of ZnS\text{ZnS} is 0.2 M0.2\text{ M} (or 0.200.20).

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