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A solution which is 10^(-3) M each in Mn...

A solution which is `10^(-3) M` each in `Mn^(2+), Fe^(2+), Zn^(2+)` and `Hg^(2+)` is treated with `10^(-16) M` sulphide ion. If `K_(sp)` of MnS, FeS, ZnS and HgS are `10^(-15), 10^(-23), 10^(-20)` and `10^(-54)` respectively, which one will precipitate first ?

A

FeS

B

MgS

C

HgS

D

ZnS

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To determine which metal sulfide will precipitate first from a solution containing `10^(-3) M` of `Mn^(2+)`, `Fe^(2+)`, `Zn^(2+)`, and `Hg^(2+)` when treated with `10^(-16) M` sulfide ions, we need to calculate the ionic product (IP) for each metal sulfide and compare it with their respective solubility product constants (Ksp). ### Step-by-Step Solution: 1. **Identify the given concentrations and Ksp values**: - Concentration of each metal ion: `10^(-3) M` - Concentration of sulfide ion: `10^(-16) M` - Ksp values: - Ksp of MnS = `10^(-15)` - Ksp of FeS = `10^(-23)` - Ksp of ZnS = `10^(-20)` - Ksp of HgS = `10^(-54)` 2. **Calculate the ionic product (IP) for each metal sulfide**: The ionic product (IP) for each metal sulfide can be calculated using the formula: \[ \text{IP} = [\text{Metal}^{2+}][\text{S}^{2-}] \] Since the concentration of metal ions is `10^(-3) M` and the concentration of sulfide ions is `10^(-16) M`, we have: \[ \text{IP} = (10^{-3})(10^{-16}) = 10^{-19} \] 3. **Compare the ionic product with Ksp values**: - For MnS: - IP = `10^(-19)` vs Ksp = `10^(-15)` (IP < Ksp, no precipitation) - For FeS: - IP = `10^(-19)` vs Ksp = `10^(-23)` (IP > Ksp, precipitation occurs) - For ZnS: - IP = `10^(-19)` vs Ksp = `10^(-20)` (IP > Ksp, precipitation occurs) - For HgS: - IP = `10^(-19)` vs Ksp = `10^(-54)` (IP > Ksp, precipitation occurs) 4. **Determine which precipitates first**: The metal sulfide with the lowest Ksp will precipitate first since it requires the least concentration of sulfide ions to exceed its Ksp. - Ksp values (lower is more soluble): - MnS: `10^(-15)` - FeS: `10^(-23)` - ZnS: `10^(-20)` - HgS: `10^(-54)` Among FeS, ZnS, and HgS, HgS has the lowest Ksp value. 5. **Conclusion**: Therefore, **HgS will precipitate first** because it has the lowest Ksp value.
To determine which metal sulfide will precipitate first from a solution containing `10^(-3) M` of `Mn^(2+)`, `Fe^(2+)`, `Zn^(2+)`, and `Hg^(2+)` when treated with `10^(-16) M` sulfide ions, we need to calculate the ionic product (IP) for each metal sulfide and compare it with their respective solubility product constants (Ksp). ### Step-by-Step Solution: 1. **Identify the given concentrations and Ksp values**: - Concentration of each metal ion: `10^(-3) M` - Concentration of sulfide ion: `10^(-16) M` - Ksp values: ...
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