Ncert Exercises & Solutions

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?

For precipitation to take place, it is required that the calculated ionic product exceeds the

K_{sp} value .

Before mixing :

[S^{2 -}] = 1.0 \times 10^{- 19} M [M^{2 +}] = 0.04 M

Valume = 10 mL Volume = 5 mL

After mixing :

\underset{Volume = (10 + 5) = 15 mL}{[S^{2 -}] = ?} \underset{Volume = 15 mL}{[M^{2 +}] = ?}

[S^{2 -}] = \frac{1.0 \times 10^{- 19} \times 10}{15} = 6.67 \times 10^{- 20} M

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

Ionic product = [M^{2 +}] [S^{2 -}]

= (1.33 \times 10^{- 2}) (6.67 \times 10^{- 20})

= 8.87 \times 10^{- 22}

This ionic product exceeds the K_{sp} of Zns and CdS . Therefore, precipitation will occur in {CdCl}_{2} and {ZnCl}_{2} solutions .

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