Question

The solubility of \[{\rm{CdS}}{{\rm{O}}_4}\] in water is \[{\rm{8}}{\rm{.0 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ mol }}{{\rm{L}}^{{\rm{ - 1}}}}\] . Its solubility in \[{\rm{0}}{\rm{.01 M }}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\] solution is _____________ \[{\rm{ \times 1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ mol }}{{\rm{L}}^{{\rm{ - 1}}}}\]. (Round off to the nearest integer.)
Assume that solubility is much less than \[{\rm{0}}{\rm{.01 M}}\]

Answer 1

Hint: A solution is a homogeneous mixture of one or more solutes in a solvent. The concentration of a substance in a saturated solution (the maximum amount of solute that can be dissolved in a known quantity of solvent) is defined as its solubility (S). Its value depends upon the nature of the solvent and temperature.

Complete Step by Step Solution:
The solubility of \[{\rm{CdS}}{{\rm{O}}_4}\]in water = \[{\rm{8}}{\rm{.0 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ mol }}{{\rm{L}}^{{\rm{ - 1}}}}\]
According to the statement, we can write

The solubility product of an electrolyte at a specific temperature may be defined as the product of the molar concentration of its ions present in a saturated solution, each concentration raised to the power equal to the number of ions produced on the dissociating one molecule of the electrolyte. The more soluble a substance is, the higher the \[{{\rm{K}}_{{\rm{sp}}}}\]value it has.
Using the given data, we can write
 \[\begin{array}{l}{\rm{S = 8 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}{{\rm{S}}^{\rm{2}}}{\rm{ = (8 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}{{\rm{)}}^{\rm{2}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = 64 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}\end{array}\]\[\]
Solubility in \[{\rm{0}}{\rm{.01M }}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{:}}\]

Due to common ion effect of \[{\rm{S}}{{\rm{O}}_{\rm{4}}}^{{\rm{2 - }}}\] solubility of \[{\rm{CdS}}{{\rm{O}}_{\rm{4}}}\] is suppressed.
\[\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}}{\rm{ = S (S + 1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{) = 64 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}\\{\rm{(S + 1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{)}} \approx {\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = S \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ = 64 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = 64 \times 1}}{{\rm{0}}^{{\rm{ - 6}}}}\end{array}\]

Hence, \[{\rm{64}}\]is the correct answer. Solubility of \[{\rm{CdS}}{{\rm{O}}_{\rm{4}}}\]in \[{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\] solution is = \[{\rm{64 \times 1}}{{\rm{0}}^{{\rm{ - 6}}}}\]

Note: Calculating solution values from equilibrium conditions and solubility constants requires careful attention to the product and reactant definitions in the equation, and a properly balanced chemical reaction, where pertinent. Forgetting that solubility is relative to other substances leads to errors in stating whether something is soluble or not. Keep in mind what orders of magnitude of concentrations will result in precipitates or not, particularly in the presence of other compounds.
Use the relevant formula and calculate precisely.