Potassium phosphate, K_{3}PO_{4} is a solid at room temperature and is soluble in water

Potassium phosphate, K_{3}PO_{4} is a solid at room temperature and is soluble in water. Which of the following statements is correct?
Select one:
a. K_{3}PO_{4} is an electrolyte.
b. K ^ + and P O 4 ^ 3- ions are solvated by water.
c. The solubility of K_{3}PO_{4} will increase with increasing temperature.
d. All of the above are correct.
13 Saline solutions (NaCl in water) used to deliver intravenous drugs are 0.89% * (w / v) . What mass of NaCl would be needed to prepare 500.0 mL of such a solution?
Select one:
Oa. 0.45 g
b. 0.89 g
Oc. 4.5 g
d. 8.9 g

The Correct Answer and Explanation is :

Question 1: Potassium phosphate solubility and properties

The correct answer is:

d. All of the above are correct.

Explanation:

• a. K₃PO₄ is an electrolyte: K₃PO₄ is an electrolyte because it dissociates into ions when it dissolves in water. In this case, it dissociates into K⁺ (potassium) ions and PO₄³⁻ (phosphate) ions. These ions conduct electricity in solution, which is the defining characteristic of an electrolyte.
• b. K⁺ and PO₄³⁻ ions are solvated by water: When K₃PO₄ dissolves in water, the K⁺ and PO₄³⁻ ions become surrounded by water molecules, a process known as solvation. The positively charged K⁺ ions are solvated by the partial negative charge of the oxygen atoms in water, while the negatively charged PO₄³⁻ ions are solvated by the partial positive charges of the hydrogen atoms in water.
• c. The solubility of K₃PO₄ will increase with increasing temperature: Like most solid solutes, the solubility of K₃PO₄ increases with temperature. As temperature rises, the kinetic energy of the molecules increases, which makes it easier for the solute (K₃PO₄) to dissolve in the solvent (water).

Therefore, all of the statements are correct, and the answer is d.

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Question 2: Saline solution mass calculation

The correct answer is:

c. 4.5 g

Explanation:
We are asked to prepare a 0.89% (w/v) NaCl solution, which means 0.89 grams of NaCl per 100 milliliters of solution.

To find the mass of NaCl needed to prepare 500.0 mL of solution, we can set up a simple calculation based on the percentage concentration:

1. Step 1: Convert the volume of solution to liters
   500.0 mL = 0.500 L.
2. Step 2: Use the formula for weight/volume percent:
   [
   \text{Mass of NaCl} = \text{Concentration} \times \text{Volume of solution}
   ]
   The concentration is 0.89% w/v, which means 0.89 grams of NaCl per 100 mL of solution. For 500.0 mL:
   [
   \text{Mass of NaCl} = 0.89 \, \text{g/100 mL} \times 500.0 \, \text{mL} = 4.45 \, \text{g}
   ]
   This rounds to 4.5 g.

Thus, the correct answer is c. 4.5 g.