Q. 9.88 - An Introduction to General, Organic, and Biological Chemistry

Question

Potassium chloride has a solubility of $43 g of KCl in 100. g$ of $H_{2}O at 50°C.$ Determine if each of the following forms an unsaturated or saturated solution at $20°C:(9.3)$

a) adding $25 g of KCl to 100. g of H_{2}O$

b) adding $15 g of KCl to 25 g of H_{2}O$

c) adding $86 g of KCl to 150. g of H_{2}O$

Step-by-Step Solution

Answer

Part a) As a corollary, this solution is unsaturated.

Part b) As a conclusion, this is a Super Saturate solution.

Part b) As a function, this is a Super Saturate solution.

Step 1: Introduction (Part a)

Saturate solution: A saturated solution is one that has the greatest levels of dissolved solute at equilibrium.  

A solution with a lower quantity than a saturated solution. It will be capable of dissolving a higher amount of solute.

Solution for super-saturation:  A solution with a stronger presence of solute solute than a saturated solution.

Step 2: Given data (Part a).

To $50°C$, $KCl$ has a solubility of roughly $\frac{43 g}{100 ml}$, or $0.43 gm{l}^{-1}$. The mixture is saturated if there is $43 g KCL in 100 ml$ of water at $50°C$.

Step 3: Explanation (Part a).

a. If $25 g KCl$ is added to $100 g$ of water, its solubility is $0.25 gm{l}^{-1}$, which is below the soluble of $KCl$ at $50°C (0.43 gm{l}^{-1})$, showing that the solution is unsaturated.

Step 4: Given data (Part b).

Evaluate the number of $KCl$ per ml in the following solution:

$\frac{15 g}{25 ml}=0.6 g m{l}^{-1}$

Step 5: Explanation (Part b).

Because this value exceeds the solubility of $KCl; 0.43 g m{l}^{-1}$, this is a Super Saturate solution.

Step 6: Given data (Part c).

Evaluate the number of  $KCl$per ml in the following solution: 

$\frac{86 g}{150 ml}=0.43 g m{l}^{-1}$

Step 7: Explanation (Part c).

So this value is higher the solubility of $KCl (0.43 g m{l}^{-1})$, this is a Super Saturate solution.