Problem 28 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

To \(20 \mathrm{ml}\) of \(0.1 \mathrm{M}-\mathrm{NaOH}\) solution, \(3 \mathrm{ml}\) of \(1 \mathrm{M}\) acetic acid solution is added. Is the solution now neutral, acidic or alkaline? How much more of the acetic acid solution we add to produce a change of \(\mathrm{pH}=0.3\) unit? \(\left(\mathrm{p} K_{\mathrm{a}}\right.\) for \(\mathrm{CH}_{3} \mathrm{COOH}\) \(=4.74, \log 2=0.3\) ) (a) acidic, \(2 \mathrm{ml}\) (b) alkaline, \(1 \mathrm{ml}\) (c) acidic, \(1 \mathrm{ml}\) (d) neutral, \(2 \mathrm{ml}\)

Step-by-Step Solution

Answer

The solution is acidic. To produce a pH change of 0.3 unit, 1 ml of 1 M acetic acid solution is needed.

Step 1: Determine Moles of NaOH and CH3COOH

First, calculate the moles of NaOH and CH3COOH by multiplying the volume in liters by the concentration in moles per liter. For NaOH: Moles of NaOH = 0.020 L * 0.1 M = 0.002 moles. For CH3COOH: Moles of CH3COOH = 0.003 L * 1 M = 0.003 moles.

Step 2: Determine Reaction Outcome

Next, determine the outcome of the reaction between NaOH and CH3COOH. Since NaOH is a strong base and CH3COOH is a weak acid, they will react in a 1:1 ratio to form CH3COONa (sodium acetate) and H2O until either the NaOH or CH3COOH is fully consumed. In this case, all 0.002 moles of NaOH will react with 0.002 moles of CH3COOH, leaving 0.001 moles of CH3COOH unreacted.

Step 3: Assess Solution Character

After the reaction, 0.001 moles of CH3COOH will remain in the solution. The presence of unreacted CH3COOH, a weak acid, makes the solution acidic. Therefore, the solution is now acidic.

Step 4: Calculate Amount of Acetic Acid Needed for pH Change

To produce a change of pH = 0.3 unit, use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]). Since the initial pH change is less than 0.3 units (assuming no significant buffer capacity), calculate the volume of 1M CH3COOH needed to decrease the pH by 0.3 units, assuming a 1:1 stoichiometry. The pH change corresponds to doubling the [HA], so you need to add the same amount of CH3COOH that's already present. This means you need to add an additional 1 ml of 1 M acetic acid solution.

Key Concepts

Understanding Acid-Base Reactions

An acid-base reaction is a chemical process where an acid and a base interact, often resulting in the formation of water (H₂O) and a salt. This interaction is a fundamental concept in chemistry that governs a broad range of phenomena, from digestion in biology to material resistance in engineering.

Essentially, these reactions can be described as a proton (H⁺) transfer process. Acids are substances that can donate protons, while bases are substances that can accept them. The strength of an acid or base is significant in determining the outcome of their reaction. For instance, a strong acid will completely dissociate in water to donate protons, while a weak acid will only partially dissociate. Similarly, a strong base will fully accept protons, whereas a weak base will only do so partially.

When a strong base, like sodium hydroxide (NaOH), reacts with a weak acid, such as acetic acid (CH₃COOH), the NaOH can neutralize equivalent moles of the weak acid, forming water and a salt, sodium acetate (CH₃COONa) in this case. In the exercise provided, 0.002 moles of NaOH react with 0.002 moles of CH₃COOH, but there are 0.003 moles of acetic acid to begin with, leaving excess acetic acid (0.001 moles unreacted) and thus, an acidic solution.

The Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation plays a pivotal role in pH calculation for buffer solutions. It is particularly useful in estimating the pH of a solution containing a weak acid and its conjugate base. The equation is expressed as:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
where:

• \(\text{pKa}\) is the acid dissociation constant
• \(\text{A}^-\) represents the concentration of the conjugate base
• \(\text{HA}\) stands for the concentration of the weak acid.

Using the equation, one can determine the effect of adding more acid or base on the pH of the solution. In the context of the exercise, by knowing the pKa of acetic acid and the reaction ratio, we can deduce the quantity of additional acetic acid required to change the pH by 0.3 units. It turns out, doubling the concentration of acetic acid [\text{HA}] will lead to the desired pH shift, indicating that 1 ml of 1 M acetic acid solution should be added.

Stoichiometry in Chemical Equations

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It allows chemists to calculate the amount of reactants needed or products formed in a reaction. The coefficients in a balanced chemical equation tell us the ratio in which substances react and are produced.

For instance, if we are given a reaction where 1 mole of NaOH reacts with 1 mole of CH₃COOH to produce 1 mole of CH₃COONa and 1 mole of H₂O, this 1:1:1:1 ratio provides the stoichiometric basis for the reaction.

Considering the provided exercise, we calculate the moles of NaOH and CH₃COOH present and use their stoichiometric ratio to determine how much of each reactant will take part in the reaction. This/basic understanding of stoichiometry is crucial for solving the problem at hand, which involves determining the remaining acidic content (CH₃COOH) after the initial reaction and then calculating the additional amount needed to change the solution's pH.