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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