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Determine the concentration of strontium ions in saturated solution of strontium sulfate, SrSO4, if the Ksp for SrSO4 is 3.2×10-7.
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Answers ( )
Answer:
[tex][Sr^{2+}]=5.66×10^{-4}M[/tex]
Explanation:
Hello there!
In this case, given the solubility equilibrium of strontium sulfate:
[tex]SrSO_4(s)\rightleftharpoons Sr^{2+}(aq)+SO_4^{2-}(aq)[/tex]
Whose equilibrium expression is:
[tex]Ksp=[Sr^{2+}][SO_4^{2-}][/tex]
In such a way, we can introduce the molar solubility, s, in the equation to obtain:
[tex]Ksp=s*s=s^2\\\\3.2×10^{-7}=s^2[/tex]
Then, we apply the square root to obtain:
[tex]s=\sqrt{3.2×10^{-7}}\\\\s=5.66×10^{-4}M[/tex]
Which is also the concentration of strontium ions:
[tex][Sr^{2+}]=5.66×10^{-4}M[/tex]
Best regards!