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  • The concentration of sulphide ion in 0. 1M HCl solution saturated with hydrogen sulphide is 1.0 × 10^–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO_4

7.73     The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10-19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO_{4},MnCl_{2},ZnCl_{2} and CdCl_{2}. in which of these solutions precipitation will take place?

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best_answer

We have,
the concentration of [S^{2-}]=1\times 10^{-19} and the volume of the solution containing sulphur ion = 10 mL.
Volume of metal salts solution added = 5mL
Before mixing,

[S^{2-}]=1\times 10^{-19}M 
[M^{2+}]=0.04 M

After mixing,

Volume = 15 mL

So, the concentration of [S^{2-}]=1\times 10^{-19}\times \frac{10}{15}
                                                    =6.67\times 10^{-20}M

concentration of [M^{2+}]=0.04 M\times \frac{5}{15}
                                          =1.33\times 10^{-2}

So, the ionic product = [M^{2+}][S^{2-}]
                                        = (6.67 \times 10^{-20})(1.33\times 10^{-2})
                                        = 8.87 \times 10^{-22}

For the precipitation of the solution, the ionic product should be greater than the corresponding K_s_p values.
K_s_p of FeS,\ MnS,\ ZnS,\ CdS are 6.3\times 10^{-18},\2.5\times 10^{-13},\ 1.6\times 10^{-24} and 8\times 10^{-27} respectively.

Hence precipitation will take place in CdCl_2 and ZnCl_2metal salts.

 

Posted by

manish

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