Respuesta :
Answer:
7×10⁻⁵ M
Explanation:
Consider the ICE take for the solubility of the solid, CaCO₃ as:
CaCO₃ ⇄ Ca²⁺ + CO₃²⁻
At t =equilibrium s s
The expression for Solubility product for CaCO₃ is:
[tex]K_{sp}=\left [ Ca^{2+} \right ]\left [ CO_3^2- \right ][/tex]
[tex]K_{sp}=s\times {s}[/tex]
[tex]K_{sp}=s^2[/tex]
Given [tex]K_{sp}=4.9\times 10^{-9}[/tex]
So, Ksp is:
[tex]4.9\times 10^{-9}=s^2[/tex]
[tex]s=\sqrt {4.9\times 10^{-9}[/tex]
s = 7×10⁻⁵ M
The concentration of Ca²⁺ is 7×10⁻⁵ M
The concentration of calcium ions [tex]Ca^{2+}(aq)[/tex] in a saturated solution of Calcium carbonate [tex]CaCO_3[/tex] is mathematically given as
s = 7*10^{-5} M
What is the concentration of Ca2+(aq) in a saturated solution of CaCO3?
Question Parameter(s):
The solubility product constant Ksp for CaCO3 is 4.9 * 10–9.
Generally, the equation for the Chemical Reaction is mathematically given as
CaCO3 ----><---- Ca^{2+} + CO3^{2-}
Thereofore
[tex]K_{sp}=\left [ Ca^{2+} \right ]\left [ CO_3^2- \right ][/tex]
Where
Ksp=4.9* 10^{-9}
Ksp=s^2
Hence
4.9* 10^{-9}=s^2
[tex]s=\sqrt {4.9* 10^{-9}[/tex]
s = 7*10^{-5} M
In conclusion, the concentration of Ca2+(aq) is
s = 7*10^{-5} M
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