A solution contains 0.100 M each of Ni2+ and Sn2+ ions. To separate them, the sulfide salt of...

Question:

A solution contains 0.100 M each of {eq}Ni^{2+} {/eq} and {eq}Sn^{2+} {/eq} ions. To separate them, the sulfide salt of each of the ions must be formed. What concentration of {eq}S^{2-} {/eq} must be added to accomplish this separation if the {eq}K_{sp} {/eq} of {eq}NiS = 6.3 \times 10^{-27} {/eq} and {eq}SnS = 1.1 \times 10^{-29} {/eq}?

Ksp:

When a solution is saturated, undissolved solutes are in equilibrium with their dissolved ions. This equilibrium state is governed by the constant known as Ksp. Like other equilibrium constants, Ksp is influenced by temperature such that the higher the temperature, the higher the Ksp. It can also be used to approximate the solubilities of sparingly soluble salts relative to each other where lower Ksp means lower solubility.

Answer and Explanation: 1

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{eq}\rm{NiS }{/eq} and {eq}\rm{SnS }{/eq} are both 1:1 binary salts. Moreover, the concentration of their cations is the same. This means, we only...

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Solubility Equilibrium: Using a Solubility Constant (Ksp) in Calculations

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Chapter 11 / Lesson 5
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Learn about solubility product constant. Understand the definition of Ksp, the Ksp formula, how to calculate Ksp, and how to find molar solubility from Ksp.


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