Chapter U6.123, Problem 3E

(a)

Interpretation Introduction

Interpretation: The three chloride salts needs to be arranged from the least soluble to the most soluble based on their solubility product, K_sp values.

Concept Introduction:The solubility product constant, K_sp of a monovalent metal chloride, MCl is given by the expression

  $\text{MCl }\left(s\right)\rightleftharpoons {\text{ M}}^{+}\left(aq\right)+{\text{ Cl}}^{-}\left(aq\right)$

Here,

  $K_{sp}=\left[M^{+}\right]\left[C{l}^{-}\right]$

The solubilities of a number of monovalent metal chlorides may be qualitatively deduced based on their K_sp values. The lower is the K_sp of the metal chloride, the lower is the solubility of the salt in water. Similarly, a chloride salt with a higher K_spvalue will have a higher solubility.

Answer to Problem 3E

The increasing order of solubility of the given chloride salts is as follows:

  $\text{AgCl }\left(s\right)<\text{CuCl }\left(s\right)<\text{TiCl }\left(s\right)$

Explanation of Solution

It has already been mentioned above that for salts having the same charges on the cations and the anions, the solubilities may be compared based on the K_sp values. A salt with a lower K_sp value will be less soluble than a salt with a higher K_sp value.

The K_sp values of the three chloride salts are in the order AgCl s_p values. Since AgCl has the lowest K_sp value, hence, AgCl will be the least soluble in water. TiCl, having the highest K_spvalue will be the most soluble in water. CuCl has intermediate K_spvalue and thus, the solubility of CuCl is intermediate to these two. Therefore, the increasing order of solubilities of the three salts are

  $\text{AgCl }\left(s\right)<\text{CuCl }\left(s\right)<\text{TiCl }\left(s\right)$

(b)

Interpretation Introduction

Interpretation: The effect of addition of NaCl on the solubility of CuCl as per the Le Chatilier principle needs to be explained.

Concept Introduction: The Le Chatilier principle states that when a stress is added to a chemical system at equilibrium, the system will adjust itself in such a way that the effect of the added stress is nullified. The solubility of a sparingly soluble salt may be altered by the addition of a common ion as per the Le Chatilier principle.

Answer to Problem 3E

The solubility of CuCl in water decreases upon the addition of NaCl due to the common ion effect.

Explanation of Solution

The ionization of CuCl in water is given by the reaction

  $\text{CuCl }\left(s\right)\rightleftharpoons {\text{ Cu}}^{2+}\left(aq\right)+{\text{ Cl}}^{-}\left(aq\right)$

The solubility product constant for the above ionization is given by the expression

  $K_{sp}=\left[{\text{Cu}}^{2+}\right]\left[{\text{Cl}}^{-}\right]$

Where the square braces, [..] denote molar concentrations at equilibrium.

NaCl ionizes completely in an aqueous solution as below.

  $\text{NaCl }\left(s\right)\to \text{ NaCl }\left(aq\right)\to {\text{ Na}}^{+}\left(aq\right)+{\text{ Cl}}^{-}\left(aq\right)$

The addition of NaCl introduces the common ion, Cl- which tends to increase [Cl^-]. Thus, K_sp tends to increase. However, K_sp, being a thermodynamic equilibrium constant, must stay constant at a particular temperature. To keep K_sp constant, [Cu⁺] must decrease. This is possible only when the solubility of CuCl in water decreases, i.e., the addition of NaCl suppresses the solubility of CuCl in water and throws off solid CuCl from the saturated solution.