Question 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: FeSO4, MnCl2, ZnCl2 and CdCl2. in which of these solutions precipitation will take place?
Class XI Equilibrium Page 230
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concept :- for precipitation, ionic product > solubility product .
here given,
[S²-] ( conc. of sulphide ion) = 10^-19 M
A/C to question,
10mL of sulphide ion is mixed with 5mL of 0.04M solution of different solute so, that final volume of solution is 15mL.
so, [S²-]mix = 10 × 10^-19/15 = 6.67 × 10^-19M
[M²+] = 5 × 0.04/15 = 1.33 × 10^-2 M
here, M²+ shows Fe²+, Mn²+, Zn²+ or Cd²+
now , ionic product of [M²+][S²-] = 1.33 × 10^-2 × 6.67 × 10^-19 = 8.87 × 10^-22 M
but Ksp of Cds and Zns is less than ionic product of [M²+][S²-]
hence , CdCl2 and ZnCl2 are precipitated as CdS and ZnS.
here given,
[S²-] ( conc. of sulphide ion) = 10^-19 M
A/C to question,
10mL of sulphide ion is mixed with 5mL of 0.04M solution of different solute so, that final volume of solution is 15mL.
so, [S²-]mix = 10 × 10^-19/15 = 6.67 × 10^-19M
[M²+] = 5 × 0.04/15 = 1.33 × 10^-2 M
here, M²+ shows Fe²+, Mn²+, Zn²+ or Cd²+
now , ionic product of [M²+][S²-] = 1.33 × 10^-2 × 6.67 × 10^-19 = 8.87 × 10^-22 M
but Ksp of Cds and Zns is less than ionic product of [M²+][S²-]
hence , CdCl2 and ZnCl2 are precipitated as CdS and ZnS.
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