A \(0.1 \mathrm{M}\) acetic acid solution is titrated against \(0.1 \mathrm{M}-\mathrm{NaOH}\) solution. What would be the difference in \(\mathrm{pH}\) between \(1 / 4\) and \(3 / 4\) stages of neutralization of the acid? (a) \(2 \log (0.75)\) (b) \(2 \log (0.25)\) (c) \(\log 3\) (d) \(2 \log 3\)

Short Answer

Expert verified
The difference in pH between the 1/4 and 3/4 stages of neutralization of the acid is (d) \(2 \log 3\).

Step by step solution

01

Understand the neutralization stages

In a titration of a weak acid with a strong base, the neutralization can be divided into stages. Here, the `1/4` stage of neutralization means that one quarter of the acetic acid has been neutralized, while the `3/4` stage indicates that three quarters of the acid has been neutralized.
02

Calculate the concentration of the acetic acid at the 1/4 stage

At the `1/4` neutralization stage, the concentration of acetic acid, which is initially 0.1 M, will become 0.075 M because one quarter (0.025 M) has reacted with NaOH.
03

Calculate the concentration of the conjugate base at the 1/4 stage

As acetic acid reacts with NaOH, it forms its conjugate base, acetate (CH3COO-). At the `1/4` stage, the concentration of acetate will be 0.025 M.
04

Write the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of the acid and the concentrations of the acid and its conjugate base: pH = pKa + log([A-]/[HA]). Here, [A-] is the concentration of acetate and [HA] is the concentration of acetic acid.
05

Calculate the pH at the 1/4 stage using the Henderson-Hasselbalch equation

Taking pKa of acetic acid as approximately 4.76, pH at the `1/4` stage can be calculated as: pH = 4.76 + log(0.025/0.075) = 4.76 - log(3).
06

Calculate the concentrations at the 3/4 stage

At the `3/4` stage, 0.075 M of acetic acid has been neutralized leaving 0.025 M acetic acid, and the concentration of acetate will be 0.075 M.
07

Calculate the pH at the 3/4 stage again using the Henderson-Hasselbalch equation

Using the Henderson-Hasselbalch equation again for the `3/4` stage, we find: pH = 4.76 + log(0.075/0.025) = 4.76 + log(3).
08

Subtract the pH at 1/4 stage from the pH at 3/4 stage to find the difference

The difference in pH is then pH at `3/4` stage - pH at `1/4` stage, which simplifies to: (4.76 + log(3)) - (4.76 - log(3)) = 2 log(3).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium refers to a state in a chemical reaction where the rates of the forward and reverse reactions are equal, resulting in no overall change in the concentration of reactants and products over time. This concept is crucial for understanding how systems behave when they reach a state of balance.

For example, when we talk about the neutralization of acetic acid with sodium hydroxide (NaOH), this reaction eventually reaches a point where the amount of acetic acid decreasing is equal to the amount of its conjugate base being formed. At any given moment in a solution at equilibrium, the ratio of the concentrations of these species remains constant. Understanding chemical equilibrium is essential for acid-base titration since it allows us to predict the outcome of a reaction under specific conditions.
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a formula used to estimate the pH of a buffer solution, based on the concentration of an acid and its conjugate base. It is expressed as:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
where \( \text{pKa} \) is the acid dissociation constant, \([\text{A}^-]\) is the concentration of the conjugate base, and \([\text{HA}]\) is the concentration of the acid. In the context of acid-base titration, this equation plays a pivotal role because it allows us to calculate the pH at any point during the titration by considering the ratio of the conjugate base to the acid.
Neutralization Reaction
A neutralization reaction occurs when an acid and a base react to form water and a salt. It is a type of chemical reaction that plays a central role in acid-base titrations. For instance, when acetic acid reacts with sodium hydroxide, the acetic acid is neutralized to form water and sodium acetate, a salt. The neutralization process can be observed through stages, as seen in our example, where the 1/4 and 3/4 stages represent how much of the acid has been converted to its conjugate base.

During titration, the pH of the solution changes as the acid is neutralized, which allows us to track the reaction's progress. By understanding the concept of neutralization, it becomes easier to grasp the changes occurring during different stages of the titration.
pH Scale
The pH scale is a measure of how acidic or basic a solution is, ranging from 0 to 14, with 7 being neutral. pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration:
\[ \text{pH} = -\log([H^+]) \]
Lower pH values correspond to higher acidity, while higher pH values indicate greater basicity. When performing a titration, such as in our exercise involving acetic acid and NaOH, the pH scale helps us understand the gradual change in acidity or basicity of the solution. By comparing pH at different stages of the titration, we can determine the equivalence point where neutralization is complete, as well as track the formation of the buffer region which helps in resisting changes in pH.

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Most popular questions from this chapter

The solubility product of \(\mathrm{AgCl}\) is \(1.0 \times 10^{-10}\). The equilibrium constant of the reaction \(\mathrm{AgCl}(\mathrm{s})+\mathrm{Br}^{-} \rightleftharpoons \mathrm{AgBr}(\mathrm{s})+\mathrm{Cl}\) is 200 and that of the reaction \(2 \mathrm{AgBr}(\mathrm{s})+\mathrm{S}^{2-} \rightleftharpoons \mathrm{Ag}_{2} \mathrm{~S}(\mathrm{~s})+2 \mathrm{Br}\) is \(1.6 \times 10^{24} .\) What is the \(K_{\mathrm{sp}}\) of \(\mathrm{Ag}_{2} \mathrm{~S}\) ? (a) \(3.2 \times 10^{16}\) (b) \(1.56 \times 10^{-49}\) (c) \(3.95 \times 10^{-25}\) (d) \(3.13 \times 10^{-17}\)

In an attempted determination of the solubility product constant of \(\mathrm{Tl}_{2} \mathrm{~S}\), the solubility of this compound in pure \(\mathrm{CO}_{2}\) free water was determined as \(2.0 \times 10^{-6} \mathrm{M}\). Assume that the dissolved sulphide hydrolyses almost completely to \(\mathrm{HS}^{-}\) and that the further hydrolysis to \(\mathrm{H}_{2} \mathrm{~S}\) can be neglected, what is the computed \(K_{\mathrm{sp}}\) ? For \(\mathrm{H}_{2} \mathrm{~S}, K_{\mathrm{al}}=1.4 \times 10^{-7}, K_{\mathrm{a} 2}=1.0 \times 10^{-14}\) (a) \(6.4 \times 10^{-23}\) (b) \(1.6 \times 10^{-23}\) (c) \(3.2 \times 10^{-17}\) (d) \(3.2 \times 10^{-24}\)

An acid type indicator, HIn differs in colour from its conjugate base \(\left(\mathrm{In}^{-}\right)\). The human eye is sensitive to colour differences only when the ratio \([\operatorname{In}] /[\mathrm{HIn}]\) is greater than 10 or smaller than \(0.1\). What should be the minimum change in the \(\mathrm{pH}\) of the solution to observe a complete colour change \(\left(K_{\mathrm{a}}=1.0 \times 10^{-5}\right)\) ? (a) \(0.0\) (b) \(1.0\) (c) \(2.0\) (d) \(5.0\)

A \(0.28 \mathrm{~g}\) sample of an unknown monoprotic organic acid is dissolved in water and titrated with a 0.1 M sodium hydroxide solution. After the addition of \(17.5 \mathrm{ml}\) of base, a pH of \(5.0\) is recorded. The equivalence point is reached when a total of \(35.0 \mathrm{ml}\) of \(\mathrm{NaOH}\) is added. The molar mass of the organic acid is (a) 160 (b) 80 (c) 40 (d) 120

Water in equilibrium with air contains \(4.4 \times 10^{-5} \% \mathrm{CO}_{2}\). The resulting carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{3}\), gives the solution a hydronium ion concentration of \(2.0\) \(\times 10^{-6} \mathrm{M}\), about 20 times greater than that of pure water. What is the \(\mathrm{pH}\) of the solution at \(298 \mathrm{~K} ?(\log 4.4=0.64\) \(\log 2=0.3\) ) (a) \(5.36\) (b) \(5.70\) (c) \(8.30\) (d) \(5.64\)

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