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Consider the molar solubility of SrCO3 in 0.10 M Sr(NO3)2 versus in pure water. Given that [tex]K_{sp}[/tex] for SrCO3 is [tex]5.4 \times 10^{-10}[/tex], which statement is true?

A. The solubility of Sr(NO3)2 increases.
B. The molar solubility of SrCO3 will be higher in 0.10 M Sr(NO3)2.
C. The molar solubility of SrCO3 will be lower in water.
D. The molar solubility of SrCO3 will be lower in 0.10 M Sr(NO3)2 because of [tex]Sr^{2+}[/tex].
E. There is not enough information given to determine an answer.

Answer :

Answer:

  • Option D(The molar solubility of [tex]SrCO_3[/tex] will be lower in 0.10 M 0 M [tex]Sr(NO_3)_2[/tex] because of [tex]Sr^{2+}[/tex]) is the correct statement

Explanation:

The equation for solubility is written as

[tex]SrCO_3(s) <===> Sr+2(aq) + CO_{3-2(aq)}[/tex]

When we initially have Sr+2 ions present in [tex]Sr(NO_3)_2[/tex] then the solubility will decrease as initially present Sr+2 ,due to common ion effect, make the reaction less favourable in forward direction and hence solubility will decrease in [tex]Sr(NO_3)_2[/tex] as compared in water where no Sr+2 ions are present initially.

Therefore,

The molar solubility of [tex]SrCO_3[/tex] will be lower in 0.10 M 0 M [tex]Sr(NO_3)_2[/tex] because of [tex]Sr^{2+}[/tex] is the correct statement

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Rewritten by : Barada

Answer:

The molar solubility of SrCO3 will be lower in 0.10 M 0 M Sr(NO3)2 because of Sr2+

Explanation:

The question reflects the phenomenon known in chemistry as common ion effect. Common ion effect refers to the decreased solubility of a solid in a solution that contains an ion in common with the solute.

In the case of our example here, both strontium carbonate and strontium nitrate possess the strontium II ion in common. The presence of this ion in strontium nitrate will decrease the solubility of strontium carbonate in accordance with Le Chateliers principle. Hence the answer.