Problem 113 - Chemistry and Chemical Reactivity

Question

A You mix \(30.0 \mathrm{mL}\) of \(0.15 \mathrm{M} \mathrm{NaOH}\) with \(30.0 \mathrm{mL}\) of \(0.15 \mathrm{M}\) acetic acid. What molecules and ions exist in this solution? List them in order of decreasing concentration.

Step-by-Step Solution

Answer

The solution contains water, acetate ions, and sodium ions in decreasing concentration.

Step 1: Determine Total Volume of Solution

When you mix the two solutions, the volumes add up. You have 30 mL of NaOH and 30 mL of acetic acid, totaling 60 mL.

Step 2: Calculate Initial Moles of Each Reactant

Calculate the moles of each reactant using the formula \( \,n = M \times V\). For NaOH: \(0.15 \, \text{mol/L} \times 30/1000 \, \text{L} = 0.0045 \, \text{mol}\). For acetic acid (CH3COOH): same calculation applies, resulting in 0.0045 mol.

Step 3: Identify Reaction Type and Write Balanced Equation

NaOH (a strong base) reacts with acetic acid (CH3COOH, a weak acid) in a neutralization reaction to form water and sodium acetate (CH3COONa). The equation is: \[ \text{NaOH} + \text{CH}_3\text{COOH} \rightarrow \text{H}_2\text{O} + \text{CH}_3\text{COONa} \]

Step 4: Determine Reaction Completion

Both reactants are in stoichiometric amounts (equimolar), so they will completely react with each other. 0.0045 mol of NaOH reacts with 0.0045 mol of CH3COOH.

Step 5: Calculate Concentration of Products

After the reaction, NaOH and CH3COOH are completely converted to CH3COONa. The concentration of CH3COONa is calculated using moles over total volume. \[ \text{Concentration of CH}_3\text{COONa} = \frac{0.0045 \, \text{mol}}{0.060 \, \text{L}} = 0.075 \, \text{M} \]

Step 6: Write Ionic Equations if Relevant

Since CH3COONa is a salt that dissolves in water, it dissociates into ions: CH3COO⁻ and Na⁺. Water remains molecular. The main species in the solution are: CH3COO⁻, Na⁺, and water.

Step 7: List All Present Species by Concentration

In decreasing order of concentration: 1. Water (solvent)2. \(\text{CH}_3\text{COO}^-\) (acetate ions, 0.075 M)3. \(\text{Na}^+\) (sodium ions, 0.075 M)

Key Concepts

Stoichiometry

Stoichiometry is the backbone of chemistry that helps us understand how substances interact in chemical reactions. It involves using balanced chemical equations to determine the proportions of reactants and products involved.
In the neutralization reaction between sodium hydroxide (NaOH) and acetic acid (CH₃COOH), stoichiometry tells us the exact amount of each reactant needed to complete the reaction.
This is crucial because the reaction only proceeds when the reactants are present in the correct mole ratio, which in this case is 1:1.
By calculating the moles of NaOH and CH₃COOH and ensuring they are equal, we confirm that the reaction goes to completion, forming water and sodium acetate (CH₃COONa).

• Use the formula for moles: \( n = M \times V \) where \( M \) is molarity, and \( V \) is volume in liters.
• In our example, converting reactants to products is 0.0045 mol for both.

The stoichiometric calculations assure us that all of the acetic acid will react completely, leaving no unused reactants.

Acetic Acid

Acetic acid, a weak acid, is notable for only partially dissociating in water.
This partial dissociation in solution means that not all acetic acid molecules release hydrogen ions (H⁺).
In the reaction with sodium hydroxide, acetic acid acts as a reactant that, in stoichiometric amounts, participates in the complete reaction to yield a salt and water.

• Unlike strong acids, acetic acid does not release all its hydrogen ions at once, contributing to its characterization as a weak acid.
• In a neutralization reaction, each mole of acetic acid reacts with a mole of NaOH to form one mole of sodium acetate.

Acetic acid's properties as a weak acid are why it is widely used in controlled reactions, allowing chemists to predict and monitor the reaction progress closely.

Sodium Acetate

Upon neutralization of acetic acid with sodium hydroxide, sodium acetate is formed. Sodium acetate is a salt that serves as a product of this acid-base reaction.
This compound readily dissolves in water, dissociating into acetate ions (CH₃COO⁻) and sodium ions (Na⁺).

• The concentration of sodium acetate in the solution is determined by the stoichiometry of the reaction, here calculated to be 0.075 M.
• Sodium acetate is often used in chemical buffers because of its ability to maintain pH levels.

The production of sodium acetate in this neutralization not only balances the equation but also exhibits how acids and bases can transform into stable products.

Ionic Equations

Ionic equations provide a closer look at the species present in the solution after the reaction. For the neutralization of acetic acid by sodium hydroxide, we consider the dissociation of the resulting products.
Sodium acetate dissociates in water to form acetate ions and sodium ions:

• Full ionic equation: \[ \text{NaOH} + \text{CH}_3\text{COOH} \rightarrow \text{H}_2\text{O} + \text{CH}_3\text{COO}^- + \text{Na}^+ \]
• Since water is a liquid and does not ionize, it remains as \(\text{H}_2\text{O}\).

Understanding this equation helps us see all the ions and compounds present after the reaction. Due to dissociation, the final solution prominently features acetate ions and sodium ions, explaining why these are listed in the concentration order. Ionic equations distill the reaction to its basic ionic movements, which is essential for grasping the chemical dynamics at play.