Problem 87
Question
When \(10 \mathrm{~mL}\) of \(0.1 \mathrm{M}\) acetic acid \((\mathrm{pKa}=5.0)\) is titrated against \(10 \mathrm{~mL}\) of \(0.1 \mathrm{M}\) ammonia solution \((\mathrm{pKb}=5.0)\) the equivalence point occurs at \(\mathrm{pH}\) (a) \(5.0\) (b) \(6.0\) (c) \(9.0\) (d) \(7.0\)
Step-by-Step Solution
Verified Answer
The pH at the equivalence point is 7.0, so the correct answer is (d) 7.0.
1Step 1: Understand the Reactions
We are dealing with a titration of a weak acid (acetic acid) and a weak base (ammonia). Both have the same concentration and balance in terms of moles because \\(\text{Moles of acetic acid} = \text{Moles of ammonia} = 0.001 \space \text{mol}\).
2Step 2: Calculate pH at Halfway Point
Initially, for weak acid and weak base titrations such as this, at the halfway point, \\(\text{pH} = \frac{\text{pKa} + \text{pKb}}{2}\).\Using \(\text{pKa} = 5.0\) and \(\text{pKb} = 5.0\), \\(\text{pH} = \frac{5.0 + 5.0}{2} = 5.0\).
3Step 3: Determine pH at Equivalence Point
At the equivalence point, the weak acid and weak base are completely neutralized, forming water and the salt of the acid and base. Since both acetic acid and ammonia are weak and have equal strength, the solution's pH will be neutral. Thus, \\(\text{pH} = 7.0\).
4Step 4: Identify the Correct Answer
Based on the calculation from Step 3, the pH at the equivalence point in this titration is \(7.0\). Therefore, the correct option is (d) \(7.0\).
Key Concepts
Equivalence Point in TitrationUnderstanding Weak Acids and Weak BasesCalculating pH in TitrationRole of Neutralization Reaction
Equivalence Point in Titration
The equivalence point in a titration is the moment during the process where the amount of acid equals the amount of base in terms of moles. The reaction is complete, and the acid and base have been completely neutralized by one another. In this specific exercise, it involves titration between a weak acid, acetic acid, and a weak base, ammonia. The equivalence point is a crucial aspect as it tells us when the titration is complete. This point can significantly vary depending on the strength of the acids and bases involved. For instance, in strong acid-strong base reactions, the equivalence point typically falls at pH 7. However, with combinations involving weak acids or weak bases, the equivalence point can differ substantially. Here, because we deal with both a weak acid and weak base with similar strength, the equivalence point results in a neutral pH of 7.
Understanding Weak Acids and Weak Bases
Weak acids, like acetic acid, and weak bases, such as ammonia, do not completely dissociate into ions when dissolved in water. This partial dissociation is what defines them as 'weak'. Their strengths are represented by their dissociation constants, pKa and pKb, respectively.
- The pKa of acetic acid is 5.0, indicating a relatively low strength as an acid.
- Similarly, the pKb for ammonia is 5.0, reflecting its low basic strength.
Calculating pH in Titration
Calculating the pH at various stages of titration is essential to understand the progress and endpoint of the process. During titration of a weak acid with a weak base, the pH can be anticipated both at the halfway point and at the equivalence point.
- At the halfway point, when equal volumes but not moles of acid and base have reacted, the pH can be found using the equation \(\text{pH} = \frac{\text{pKa} + \text{pKb}}{2}\).
- Using the exercise's values, this results in a pH of 5.0 halfway.
Role of Neutralization Reaction
Neutralization is the chemical reaction forming the backbone of titration between acids and bases. It involves the hydrogen ions from the acid reacting with the hydroxide ions from the base to form water. In a weak acid and weak base setup like our exercise, after neutralization, a salt is formed. Here, acetic acid and ammonia neutralize to form ammonium acetate and water.
However, because both the acid and base are weak, the resulting solution at the equivalence point doesn’t create excess hydrogen ions or hydroxide ions that would significantly affect the pH.
This neutralization supports the formation of a neutral pH 7 at the equivalence point. Understanding this process helps grasp why weak acid and weak base titrations have different pH characteristics than those involving strong acids and bases.
Other exercises in this chapter
Problem 85
At \(25^{\circ} \mathrm{C}\) the \(\mathrm{pH}\) of solution containing \(0.10 \mathrm{M}\) sodium acetate and \(0.03 \mathrm{M}\) acetic acid is \([\mathrm{pKa
View solution Problem 86
At \(80^{\circ} \mathrm{C}\), distilled water \(\left(\mathrm{H}_{3} \mathrm{O}^{+}\right)\)concentration is equal to \(1 \times 10^{-6} \mathrm{~mol} /\) litre
View solution Problem 88
\(40 \mathrm{ml}\) of \(0.1 \mathrm{M}\) ammonia solution is mixed with 20 mol of \(0.1 \mathrm{M} \mathrm{HCl}\). What is the \(\mathrm{pH}\) of the mixture? (
View solution Problem 89
The solubility of \(\mathrm{AgCl}\) in \(0.2 \mathrm{M} \mathrm{NaCl}\) is \([\mathrm{Ksp} \mathrm{AgCl}=\) \(\left.1.8 \times 10^{-10}\right]\) (a) \(1.8 \time
View solution