Respuesta :
Answer : The correct answer is [tex]5.3\times 10^{-4}moles/L[/tex].
Solution : Given,
[tex]K_{sp}=2.8\times 10^{-7}[/tex]
The balanced equilibrium reaction is,
[tex]SrSO_4\rightleftharpoons Sr^{2+}+SO^{2-}_4[/tex]
At equilibrium s s
The expression for solubility constant is,
[tex]K_{sp}=[Sr^{2+}][SO^{2-}_4][/tex]
Now put the given values in this expression, we get
[tex]2.8\times 10^{-7}=(s)(s)\\2.8\times 10^{-7}=s^2\\s=5.29\times 10^{-4}=5.3\times 10^{-4}moles/L[/tex]
Therefore, the solubility of [tex]SrSO_4[/tex] in moles/L is [tex]5.3\times 10^{-4}[/tex].
Answer: The solubility of [tex]SrSO_4[/tex] is [tex]5.3\times 10^{-4}mol/L[/tex]
Explanation:
It is given that [tex]K_{sp}[/tex] of strontium sulfate is [tex]2.8\times 10^{-7} [/tex]
The balanced equilibrium reaction for ionization of [tex]SrSO_4[/tex] is given by:
[tex]SrSO_4\rightleftharpoons Sr^{2+}+SO^{2-}_4[/tex]
At equilibrium: s s
The equation to calculate solubility constant is given as:
[tex]K_{sp}=[Sr^{2+}][SO^{2-}_4][/tex]
Now put the given values in above equation, we get:
[tex]2.8\times 10^{-7}=(s)(s)\\2.8\times 10^{-7}=s^2\\s=5.29\times 10^{-4}\approx 5.3\times 10^{-4}moles/L[/tex]
Therefore, the solubility of [tex]SrSO_4[/tex] is [tex]5.3\times 10^{-4}mol/L[/tex]
