Question

Salt solubilities can be compared by the concentration of cation formed when the salt dissolves in the general reaction: MaXb(s) ⇌ a Mb+(aq) + b Xa-(aq). Given the following salts and their equilibrium constants for the reaction above at 25°C, which salt is the least soluble?

Answer 1

Answer:

The correct answer is C

Explanation:

When looking at Kc values, the smaller the value of Kc, the smaller the solubility of the compound.

Thus, if we want to find the compound that is least soluble, we simply need to identify the salt with the smallest Kc.

In this case, SrCO3 has the smallest Kc (5.6*10^-10), so it follows that of the four choices, SrCO3 is the least soluble.

Answer 2

Question:

A) MgCO₃, Kc = 6.8 × 10⁻⁶

B) BaCO₃, Kc = 2.6 × 10⁻⁹

C) SrCO₃, Kc = 5.6 × 10⁻¹⁰

D) CaCO₃, Kc = 5.0 × 10⁻⁹

Answer:

The correct option is;

C) SrCO₃, Kc = 5.6 × 10⁻¹⁰

Explanation:

To solve the question, we note that by analyzing each salt as follows

MgCO₃, Kc = 6.8 × 10⁻⁶

MgCO₃  ⇄ Mg + CO₃

Here the kc is given as

$k_c =\frac{[Mg][CO_3]}{[MgCO_3]}$

and the Ksp can be represented as  

$k_{sp}$ = [Mg][CO₃]

Therefore, for amount of salt we have the solubility product $k_{sp}$, proportional to kc

Therefore $k_{sp}$ is directly proportional to kc since the salt are 1:1, that is the amount ratio of the dissociated ions, (positive to negative ions concentration ratio) is 1:1

Therefore from the given kc values, we have

MgCO₃, Kc = 6.8 × 10⁻⁶ > CaCO₃, Kc = 5.0 × 10⁻⁹ > BaCO₃, Kc = 2.6 × 10⁻⁹ > SrCO₃, Kc = 5.6 × 10⁻¹⁰

Which gives the salt with the least solubility as, SrCO₃, with Kc = 5.6 × 10⁻¹⁰.