In presence of HCl - H 2 S results the precipitation of Group -2 elements but no Gp -4 elements during quantitative analysis- It is due toA- Higher concentration of S 2-B- Higher concentration of H -C- Lower concentration of S 2-D- Lower concentration of H -

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Standard XII

Chemistry

Crystal Field Split in Octahedral Complexes

In presence o...

Question

In presence of HCl, $H_{2} S$ results the precipitation of Group – 2 elements but no Gp – 4 elements during quantitative analysis. It is due to

Higher concentration of S^2-

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Higher concentration of H⁺

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Lower concentration of S^2-

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Lower concentration of H⁺

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Solution

The correct option is C
Lower concentration of S^2-

In HCl, the dissociation $H_{2} S$ is suppressed. Hence it will have lower concentration of $S^{2 -}$ ions, which results in the precipatation of those sulphides having very low solubility product.

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In presence of HCl, H2S results the precipitation of Group – 2 elements but no Gp – 4 elements during quantitative analysis. It is due to

The correct option is C Lower concentration of S2- In HCl, the dissociation H2S is suppressed. Hence it will have lower concentration of S2− ions, which results in the precipatation of those sulphides having very low solubility product.

In presence of HCl, $H_{2} S$ results the precipitation of Group – 2 elements but no Gp – 4 elements during quantitative analysis. It is due to

The correct option is C
Lower concentration of S^2-

In HCl, the dissociation $H_{2} S$ is suppressed. Hence it will have lower concentration of $S^{2 -}$ ions, which results in the precipatation of those sulphides having very low solubility product.