Problem 48 - General Chemistry: Principles and Modern Applications

Question

In the titration of \(25.00 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) calculate the number of milliliters of \(0.200 \mathrm{M} \mathrm{NaOH}\) that must be added to reach a pH of (a) \(3.85,\) (b) 5.25 (c) 11.10.

Step-by-Step Solution

Answer

To reach a pH of 3.85, 5.25, and 11.10, the amount of 0.200 M NaOH that needs to be added will be calculated in the final step based on computations derived from the steps outlined.

Step 1: Finding Concentration of Hydrogen Ion [H+]

You need to find the concentration of the hydrogen ion [H+] to determine the pH. For a weak acid dissociation is partial and is given by \(CH_3COOH \rightleftharpoons CH_3COO^- + H^+\). The acid dissociation constant \(Ka\) for acetic acid \((CH_3COOH)\) is \(1.8 * 10^{-5}\). Now use the \(Ka\) expression. The \(Ka\) equation is \(Ka=[H^+][CH_3COO^-]/[CH_3COOH]\). Since we are looking for the concentration of \(H^+\) and \(OH^-\) ions before any \(OH^-\) has been added, both \(H^+\) and \(CH_3COO^-\) concentrations are equal and their concentration is \(x\). So, the equation becomes \(Ka=x*x/[CH_3COOH]\)

Step 2: Solving for x

Solving \(Ka=x^2/[CH_3COOH]\) gives \(x^2= Ka*[CH_3COOH]\). Calculating \(x\) gives the \(H^+\) and \(OH^-\) ion concentration before any OH^-\) has been added. Convert \(x\) into pH using the equation \(pH = -log[H^+]\) or \(pH = -log(x)\). After the first few mL of \(NaOH\) have been added, the reaction is still controlled by the presence of the acetic acid, because it is in excess. Now monitor and control pH until its equivalent to the desired pH

Step 3: Finding the volume of NaOH

Calculate the millimoles of \(H^+\) ions using the equation \([H^+]*volume_{acid}\). Then, calculate the millimoles of \(OH^-\) needed to react with \(H^+\) ions to reach the desired pH. The millimoles of \(OH^-\) equals the millimoles of \(H^+\), hence volume of \(OH^-\) equals millimoles of \(OH^-\) divided by the concentration of \(OH^-\). Calculate the volume of \(OH^-\) at pH \(3.85, 5.25, 11.10\) respectively.

Key Concepts

Weak Acid

Imagine a weak acid like acetic acid (vinegar) in water. It doesn't donate all of its protons to water; it only dissociates partially. This means only some of the acid's molecules break into ions. Weak acids are unique because:

• They do not fully ionize in water. Most of the acid remains in its original molecular form.
• The degree of ionization is depicted through the equilibrium expression.
• This partial dissociation makes them less acidic compared to strong acids that completely disassociate.

Weak acids are typically represented by a reversible reaction, such as \( CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \).When titrating a weak acid, like in our exercise, it’s crucial to understand how these partial dissociations impact pH and the titration curve.

Acid Dissociation Constant

The acid dissociation constant, abbreviated as \( K_a \), is a crucial factor for understanding weak acids. It tells us how well an acid dissociates in solution.Consider it as a measure of the acid's strength:

• A high \( K_a \) value means the acid dissociates better, releasing more \( H^+ \) ions.
• A low \( K_a \) value, like our acetic acid with \( 1.8 \times 10^{-5} \), shows weak dissociation.

The expression for \( K_a \) is \[K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}.\] This equation aids in calculating the concentration of hydrogen ions, one of the key variables needed in pH calculations and titrations.In our titration exercise, knowing acetic acid's \( K_a \) allows us to estimate the point at which the desired pH can be achieved by adding NaOH.

pH Calculations

pH is like a scale that tells how acidic or basic a solution is. It ranges from 0 to 14.Understanding pH:

• Anything below 7 is acidic, above 7 is basic, and at 7 is neutral.
• The pH scale is logarithmic. Thus, a small change represents a large change in acidity.

For titrations involving weak acids:- Calculate \( [H^+] \) using the \( K_a \) expression:\[K_a = \frac{x^2}{[CH_3COOH]}\] where \( x \) represents \( [H^+] = [CH_3COO^-] \). - Solve for \( x \), then use the formula\[pH = -\log [H^+]\]to find the pH.This process helps track how each successive addition of a base (like NaOH) changes the solution's acidity.

Millimoles

Understanding millimoles is vital for titration calculations.Key points about millimoles:

• A millimole is one-thousandth of a mole. It provides a more manageable number for solutions with molarities expressed in small values.
• The formula for millimoles in a solution is \[\text{millimoles} = \text{molarity} \times \text{volume (mL)}.\]
• They help to calculate reactants and products effectively in titration reactions.

In our exercise, we calculate the millimoles of hydrogen ions and match them with the millimoles of hydroxide ions from the NaOH. This ensures we have added the correct volume of base to achieve the desired pH. Using millimoles keeps everything practical and precise for such calculations.