Reaction Equation for CaCl2(aq) NH3(aq) in Aqueous Solution

Understanding the Interaction Between CaCl2(aq) and NH3(aq): A Comprehensive Guide

When considering the interaction between calcium chloride (CaCl2) and ammonia (NH3) in aqueous solution, it is crucial to understand the individual behaviors of these substances in water and how their ions can interact.

Step-by-Step Breakdown of the Reaction

Dissociation of CaCl2 in Water: Calcium chloride (CaCl2) dissociates into calcium ions (Ca2 ) and chloride ions (Cl-).

CaCl2(aq) → Ca2 (aq) 2 Cl-(aq)

Behavior of Ammonia in Water: Ammonia (NH3) behaves as a weak base, reacting with water to form ammonium ions (NH4 ) and hydroxide ions (OH-).

NH3(aq) H2O(l) ? NH4 (aq) OH-(aq)

When CaCl2 and NH3 are mixed in solution, there is no direct chemical interaction producing a new compound. Instead, the ions remain in solution, and the equilibrium of ammonia with water must be considered.

CaCl2(aq) NH3(aq) H2O(l) ? Ca2 (aq) 2 Cl-(aq) NH4 (aq) OH-(aq)

While there is no direct reaction producing a new product, the presence of NH3 can affect the equilibrium in the solution.

Formation of Precipitate Ca(OH)2

However, under certain conditions, a precipitate of calcium hydroxide (Ca(OH)2) can form. This happens if the concentration of OH- ions is high enough to exceed the solubility product (Ksp) of Ca(OH)2.

Ksp(Ca(OH)2) [Ca2 ][OH-]^2 7.9 x 10^-6

The solubility equilibrium can be determined by considering the base dissociation constant (Kb) of ammonia (NH3).

Kb(NH3) [NH4 ][OH-]/[NH3] 1.8 x 10^-5

Let's consider specific concentrations to illustrate this:

Example Scenario:

Suppose we have 0.10 M CaCl2 solution. Since CaCl2 completely dissociates, [Ca2 ] 0.10 M. Further, suppose we have a 6.0 M NH3 solution.

NH3 H2O ? NH4 OH-

Initial Concentrations:

NH3: 6.0 M NH4 : 0 M OH-: ~0 M

Change in Concentrations:

-x M in NH3 x M in NH4 x M in OH-

Equilibrium Concentrations:

NH3: 6-x M NH4 : x M OH-: x M

From the base dissociation constant (Kb), we can solve for x:

[x][x]/[6-x] 1.8 x 10^-5

Solving for x, we get x 0.010 M.

Let's check the ion product [Ca2 ][OH-]^2:

[Ca2 ][OH-]^2 [0.10][0.010]^2 1.0 x 10^-5

Since this exceeds the Ksp of Ca(OH)2 (7.9 x 10^-6), a precipitate of Ca(OH)2 will form.

CaCl2(aq) NH4OH(aq) ? Ca(OH)2(s) CaCl2(s)

Note: NH4OH(aq) represents the ions formed from the equilibrium ionization of NH3.

Conclusion

In summary, while there is no direct reaction producing a new product from CaCl2 and NH3, the presence of NH3 can affect the equilibrium in the solution and potentially lead to the formation of a precipitate of Ca(OH)2 if the concentration of OH- ions is high enough.

It would be straightforward to test this in the lab by observing what happens when these solutions are mixed.