(S) (s) (5) A 2.496 gram sample of KPO4 is dissolved in a 250.0mL stock solution. a. Determine the molarity of the whole solution. M=moles Volume L 2.4969|Imol 2129 0.0 b. Determine the molarity of the K* ions. c. If 10mL of the stock solution is then removed and flask, what is the concentration of the diluted solu

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**Educational Content: Calculating Molarity and Net Ionic Equations**

**A. Problem Overview:**

A 2.496 gram sample of K₃PO₄ is dissolved in a 250.0 mL volumetric flask to make a stock solution.

**Steps to Solve:**

**a. Determine the Molarity of the Whole Solution:**

1. Calculate moles of K₃PO₄:
   - Mass: 2.496 g
   - Molar mass of K₃PO₄: (39 * 3) + 31 + (16 * 4) = 212 g/mol
   - Moles of K₃PO₄ = 2.496 g / 212 g/mol = 0.01177 mol

2. Calculate molarity (M) of K₃PO₄:
   - Volume = 250.0 mL = 0.250 L
   - M = moles / volume = 0.01177 mol / 0.250 L = 0.04708 M

\[ \text{Calculated molarity is boxed as } M = 0.04708 \, \text{M} \]

**b. Determine the Molarity of the K⁺ Ions:**

K₃PO₄ dissociates into 3 K⁺ ions. Therefore, the molarity of K⁺ = 3 * 0.04708 M.

**c. Concentration after Dilution:**

If 10 mL of the stock solution is removed and placed in a 100 mL volumetric flask, calculate the concentration of the diluted solution using the formula:

\[ C_1V_1 = C_2V_2 \]
\[ C_2 = \frac{C_1 \times V_1}{V_2} \]

**d. Net Ionic Equation:**

Example reaction:
\[ \text{Ni}^{2+} + 2 \text{e}^- \rightarrow \text{NiS (s)} \]

**Reality Check:** Make sure your Net Ionic Equation supports your driving force.
Transcribed Image Text:**Educational Content: Calculating Molarity and Net Ionic Equations** **A. Problem Overview:** A 2.496 gram sample of K₃PO₄ is dissolved in a 250.0 mL volumetric flask to make a stock solution. **Steps to Solve:** **a. Determine the Molarity of the Whole Solution:** 1. Calculate moles of K₃PO₄: - Mass: 2.496 g - Molar mass of K₃PO₄: (39 * 3) + 31 + (16 * 4) = 212 g/mol - Moles of K₃PO₄ = 2.496 g / 212 g/mol = 0.01177 mol 2. Calculate molarity (M) of K₃PO₄: - Volume = 250.0 mL = 0.250 L - M = moles / volume = 0.01177 mol / 0.250 L = 0.04708 M \[ \text{Calculated molarity is boxed as } M = 0.04708 \, \text{M} \] **b. Determine the Molarity of the K⁺ Ions:** K₃PO₄ dissociates into 3 K⁺ ions. Therefore, the molarity of K⁺ = 3 * 0.04708 M. **c. Concentration after Dilution:** If 10 mL of the stock solution is removed and placed in a 100 mL volumetric flask, calculate the concentration of the diluted solution using the formula: \[ C_1V_1 = C_2V_2 \] \[ C_2 = \frac{C_1 \times V_1}{V_2} \] **d. Net Ionic Equation:** Example reaction: \[ \text{Ni}^{2+} + 2 \text{e}^- \rightarrow \text{NiS (s)} \] **Reality Check:** Make sure your Net Ionic Equation supports your driving force.
Expert Solution
Step 1

a. Molariy of a solution = weight of the solute x 1000/( molar mass of the solute*volume of the solution)

M= 2.496*1000/(212*250) = 0.0471 M

b. 1.0 mole of K3PO4  contains 3.0 moles of K+ ions.

The molarity of K+ ions is 3*9.0471 = 0.141 M

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