(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
(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
Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F14c6963b-cbf2-45b0-9f60-9e3d334b54c7%2F3d4ed025-973d-43df-b1fa-d86519467d4e%2Fixp2nhwd.jpeg&w=3840&q=75)
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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