Q. 9.115 - An Introduction to General, Organic, and Biological Chemistry

Question

Potassium fluoride has a solubility of $92 g of KF$ in $100 g of H_{2}O$ at ${18}^{\circ }C$. Determine if each of the following mixtures forms an unsaturated or saturated solution at ${18}^{\circ }C.$

a. adding $35 g of KF$ to $25 g of H_{2}O$

b. adding $42 g of KF$ to $50 g of H_{2}O$

c. adding $145 g of KF$ to $150 g of H_{2}O$

Step-by-Step Solution

Answer

(Part a) Adding $35 g of KF$ to $25 g of H_{2}O$create saturated solution.

(Part b) Adding $42 g of KF$ to $50 g of H_{2}O$create unsaturated solution.

(Part c) Adding $145 g of KF$ to $150 g of H_{2}O$create saturated solution.

Step1: Introduction.(Part a)

Solution that is saturated:

The solution has reached saturation until no more solute could be added to the solvent.

Solution with no saturation:

The solution is said to be unsaturated whenever there is a tendency for more solute to dissipate in the solvent.

Step2: Given Information (part a).

The following data was available to establish the solution level:

Adding $35 g of KF$ in $25 g of H_{2}O$ solution.

Step3: Explanation (part a).

Evaluate the conversion factor as follows:

$$\begin{gathered} =25 \mathrm{g}\text{ of }{\mathrm{H}}_2\mathrm{O}\times \frac{92 \mathrm{g} \mathrm{KF}}{100 \mathrm{g}\text{ of }{\mathrm{H}}_2\mathrm{O}} \\ =23 g KF \end{gathered}$$

As a consequence, the maximum is $23 g KF$ and the given is $35 g KF$. Hence, the solution is saturated.

Step4: Given Information (part b).

The following data was available to establish the solution level:

Adding $42 g of KF$ in$50 g of H_{2}O$

Step5: Explanation (part b).

Evaluate the conversion factor as follows:

$$\begin{gathered} =50 \mathrm{g}\text{ of }{\mathrm{H}}_2\mathrm{O}\times \frac{92 \mathrm{g} \mathrm{KF}}{100 \mathrm{g}\text{ of }{\mathrm{H}}_2\mathrm{O}} \\ =46 g KF \end{gathered}$$

As a consequence, the maximum is $46 g KF$and the given is $45 g KF$. Hence, the solution is unsaturated.

Step6: Given Information (part c).

The following data was available to establish the solution level:

Adding $145 g of KF$ in$150 g of H_{2}O$

Step7: Explanation (part c).

Evaluate the conversion factor as follows:

$$\begin{gathered} =150 \mathrm{g}\text{ of }{\mathrm{H}}_2\mathrm{O}\times \frac{92 \mathrm{g} \mathrm{KF}}{100 \mathrm{g}\text{ of }{\mathrm{H}}_2\mathrm{O}} \\ =138 g KF \end{gathered}$$

As a consequence, the maximum is $138 g KF$ and the given is $150 g KF$. Hence, the solution is saturated.