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Chapter 15: Problem 111

How many moles of each ion are present in \(11.7 \mathrm{~mL}\) of \(0.102 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution?

Short Answer

Expert verified
In the $11.7 \mathrm{~mL}$ of $0.102 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}$ solution, there are 0.0035802 moles of Na鈦� and 0.0011934 moles of PO鈧劼斥伝 ions.

Step by step solution

01

Determine the dissociation of Na鈧働O鈧�

When sodium phosphate dissolves in water, it dissociates into its constituent ions: \( Na_{3}PO_{4} \rightarrow 3Na^{+} + PO^{3-}_{4} \) This tells us that one mole of sodium phosphate will yield three moles of sodium ions (Na鈦�) and one mole of phosphate ions (PO鈧劼斥伝) upon dissociation.
02

Convert volume to liters

We need to express the volume of the given solution in liters. Given the volume of 11.7 mL, we can convert this to liters by dividing it by 1000: \( 11.7 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}} = 0.0117 \mathrm{~L} \)
03

Calculate moles of Na鈧働O鈧�

Now, we'll use the given molarity and the volume in liters to determine the number of moles of sodium phosphate in the solution: Moles of Na鈧働O鈧� = Molarity 脳 Volume Moles of Na鈧働O鈧� = 0.102 M 脳 0.0117 L = 0.0011934 moles
04

Calculate moles of each ion

Based on the stoichiometry of the dissociation reaction discussed in Step 1, we can now calculate the moles of each ion: Moles of Na鈦� = 3 脳 Moles of Na鈧働O鈧� Moles of Na鈦� = 3 脳 0.0011934 moles = 0.0035802 moles Moles of PO鈧劼斥伝 = 1 脳 Moles of Na鈧働O鈧� Moles of PO鈧劼斥伝 = 1 脳 0.0011934 moles = 0.0011934 moles In the given sodium phosphate solution, there are 0.0035802 moles of Na鈦� and 0.0011934 moles of PO鈧劼斥伝 ions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sodium Phosphate Dissociation
When sodium phosphate ( Na_3PO_4 ) is placed into water, it undergoes a process called dissociation. Understanding this is key in solution chemistry. When dissolved, sodium phosphate separates into its constituent ions:
  • Three sodium ions ( Na^+ )
  • One phosphate ion ( PO_4^{3-} )
Each molecule of Na_3PO_4 yields these ions, making the reaction important for various calculations, especially in ion concentration and stoichiometry.
Stoichiometry
Stoichiometry is the method used to calculate the amounts of reactants and products in chemical reactions.In the dissociation of sodium phosphate, stoichiometry helps us understand how the ions are produced. The dissociation equation:\[ Na_3PO_4 \rightarrow 3Na^+ + PO_4^{3-} \]shows that for every 1 mole of Na_3PO_4dissolved, 3 moles of Na^+and 1 mole of PO_4^{3-}are produced.This ratio is crucial for calculating the concentrations and amounts of the ions in a given solution.
Ion Concentration Calculation
Calculating the concentration of ions in a solution is essential for understanding chemical behavior in solutions.To find the moles of ions from a solution:1. Convert the volume of the solution to liters (11.7 mL = 0.0117 L in this case).2. Use the molarity (M) and volume (L) to find the moles of Na_3PO_4:\[ \text{Moles of } Na_3PO_4 = 0.102 \times 0.0117 = 0.0011934 \, \text{moles}\]3. Apply the stoichiometry from the dissociation to find:
  • Moles of Na^+ = 3 \times 0.0011934 = 0.0035802 \, \text{moles}
  • Moles of PO_4^{3-} = 0.0011934 \, \text{moles}
The ion concentration calculation leverages stoichiometry to determine the number of ions in a given volume of solution.
Solution Chemistry
Solution chemistry studies how substances dissolve and behave in solutions, focusing on the solvent and solute interactions. Sodium phosphate dissociation in water is an example where solution chemistry concepts apply. Understanding properties like concentration, molarity, and the nature of solute-solvent interactions is fundamental:
  • Molarity: refers to the number of moles of solute per liter of solution, a key to calculating concentrations.
  • Dissolution: breaking down of substances into ions, like Na_3PO_4 dissolving into Na^+ and PO_4^{3-} .
Grasping solution chemistry concepts is crucial for predicting how solutions will react or what properties they might exhibit, particularly concerning ion concentration and reactivity.

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Most popular questions from this chapter

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