The solubility of two slightly soluble salts of M^2+ , MA and MZ_2, are the same, 4 x 10^â4 mol/L.

Chemistry

1 answer:

LekaFEV [45]3 years ago

7 0

Answer:

a) MZ₂

b) They have the same concentration

c) 4x10⁻⁴ mol/L

Explanation:

a) The solubility (S) is the concentration of the salt that will be dissociated and form the ions in the solution, the solubility product constant (Kps) is the multiplication of the concentration of the ions elevated at their coefficients. The concentration of the ions depends on the stoichiometry and will be equivalent to S.

The salts solubilization reactions and their Kps values are:

MA(s) ⇄ M⁺²(aq) + A⁻²(aq) Kps = S*S = S²

MZ₂(aq) ⇄ M⁺²(aq) + 2Z⁻(aq) Kps = S*S² = S³

Thus, the Kps of MZ₂ has a larger value.

b) A saturated solution is a solution that has the maximum amount of salt dissolved, so, the concentration dissolved is solubility. As we can notice from the reactions, the concentration of M⁺² is the same for both salts.

c) The equilibrium will be not modified because the salts have the same solubility. So, let's suppose that the volume of each one is 1 L, so the number of moles of the cation in each one is 4x10⁻⁴ mol. The total number of moles is 8x10⁻⁴ mol, and the concentration is:

8x10⁻⁴ mol/2 L = 4x10⁻⁴ mol/L.

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2. What ions are present in what ratio in a solution of aqueous calcium chloride?

Alenkasestr [34]

Answer:

$\mathrm{Ca}^{2+} \text { and } \mathrm{Cl} \text { - ions are present in } 1: 2 \text { ratio in a solution of aqueous calcium chloride. }$

Explanation:

Here in Calcium Chloride ionic bond is present in between calcium and chlorine atoms. As we know according to Octet rule calcium have two excess atoms and for matching nearest noble gas electronic configuration. It donate two electrons to gain more stability and form $\mathrm{Ca}^{2+}$ , while chlorine is deficient from one electron to meet nearest noble gas electronic configuration therefore two chlorine atoms accept excess electron from calcium individually and form two $\mathrm{Cl}^{-}$ ions.

$\text { Equation is as follows: } \mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-} \rightarrow \mathrm{CaCl}_{2}$

Hence aqueous solution of calcium chloride breaks the ionic bond pairing in one $\mathrm{Ca}^{2+}$ and two $\mathrm{Cl}^{-}$ ions: $\mathrm{CaCl}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \quad \mathrm{Ca}^{2+}(\mathrm{ag})+2 \mathrm{Cl}(\mathrm{ag})$