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Chapter 12: Problem 421

\(100 \mathrm{ml}\) of \(0.1 \mathrm{M} \mathrm{NaOH}\) is mixed with \(100 \mathrm{ml}\) of \(0.15 \mathrm{M}\) HOAC. \(K_{a}=1.75 \times 10^{-5}\) for acetic acid, what is the \(\mathrm{pH}\) of this mixture?

Short Answer

Expert verified
The pH of the mixture can be calculated using the Henderson-Hasselbalch equation: \( \mathrm{pH} = \mathrm{pKa} + \log \left( \cfrac{[\mathrm{Base}]}{[\mathrm{Acid}]} \right) \). The initial moles of NaOH and HOAC are 0.01 and 0.015, respectively. After reaction, the excess acetic acid is 0.005 moles. The concentrations of HOAC and OAc- are 0.025 M and 0.05 M, respectively. The pKa of acetic acid is 4.756. Substituting the values, we get the pH of the mixture to be approximately 5.06.

Step by step solution

01

Calculate Initial Amounts of Acid and Base

Firstly, calculate the initial number of moles of NaOH and HOAC (molarity * volume). The number of moles of NaOH = \(0.1 \, \mathrm{M} \times 0.1 \, \mathrm{l} = 0.01 \, \mathrm{moles}\) The number of moles of HOAC = \(0.15 \, \mathrm{M} \times 0.1 \, \mathrm{l} = 0.015 \, \mathrm{moles}\)
02

Determine the Reaction And the Excess Component

The reaction that occurs is \( \mathrm{HOAC} + \mathrm{OH^{-}} \rightarrow \mathrm{H}_{2}\mathrm{O} + \mathrm{OAc^{-}}\). Sodium hydroxide is a strong base that reacts completely with acetic acid, a weak acid. Hence after the reaction, we will have an excess of acetic acid since we started with more moles of it. The excess acetic acid will be \(0.015 - 0.01 = 0.005 \, \mathrm{moles}\).
03

Calculate the Concentration of the Acid and its Conjugate Base

After the reaction, we will have 0.005 moles of HOAC and 0.01 moles of OAc-. The total volume of the solution is 200ml or 0.2l. So the concentration of HOAC or [HOAC] = \(0.005 / 0.2 = 0.025 \, \mathrm{M}\) and concentration of OAc- or [OAc-] = \(0.01 / 0.2 = 0.05 \, \mathrm{M}\).
04

Calculate the pH

Now, apply the Henderson-Hasselbalch equation which is \( \mathrm{pH} = \mathrm{pKa} + \log \left( \cfrac{[\mathrm{Base}]}{[\mathrm{Acid}]} \right) \), where pKa = -log(Ka). Here, Base refers to the conjugate base \( \mathrm{OAc^{-}} \) and Acid refers to \( \mathrm{HOAC} \). Firstly, calculate the pKa = -\(\log (1.75 \times 10^{-5}) = 4.756 \). Then, substitute values into the Henderson-Hasselbalch equation: \( \mathrm{pH} = 4.756 + \log \left( \cfrac{0.05}{0.025} \right ) = 4.756 + 0.301 = 5.057 \). This means the pH of this mixture is approximately 5.06.

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Key Concepts

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

Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a mathematical expression that relates the pH of a buffer solution to the concentration of its acid-base components. It's indispensable when studying acid-base titration as it simplifies the pH calculation process for buffer solutions. The equation is written as:

\[\begin{equation} pH = pKa + \text{log} \bigg(\frac{{[\text{Base}]}}{{[\text{Acid}]}}\bigg)\end{equation}\]

Where pKa is the negative logarithm of the acid dissociation constant (Ka), [Base] is the concentration of the conjugate base, and [Acid] is the concentration of the weak acid. To apply the Henderson-Hasselbalch equation effectively, you must first comprehend the acid-base reaction involved and accurately calculate the concentrations of the weak acid and its conjugate base after the reaction has taken place. The beauty of this equation lies in its ability to approximate the pH of a solution without requiring complex equilibrium calculations 鈥� a real time saver for students and chemists alike.

Acid-Base Titration
Acid-base titration is a quantitative analytical method used to determine the concentration of an unknown acid or base solution by reacting it with a solution of known concentration, referred to as the titrant. During a titration, the point at which the reactants are perfectly stoichiometrically neutralized is known as the equivalence point. Volumetric data collected from the titration process help in calculating the concentration of the unknown solution.

The process involves carefully adding the titrant to a flask containing the unknown solution and an appropriate pH indicator or using a pH meter to monitor the pH change. A sharp change in pH indicates the equivalence point has been reached. Understanding the stoichiometry of the acid-base reaction is crucial, as is having a grasp of the concept of conjugate acid-base pairs which play a key role in determining the pH at various stages of the titration.

Conjugate Acid-Base Pairs
Conjugate acid-base pairs are two species that differ by the presence or absence of a proton (H鈦�). In any acid-base reaction, the acid donates a proton to the base, resulting in the formation of a conjugate base and a conjugate acid, respectively. The pair represents two sides of a chemical equilibrium process.

For example, in the solution of acetic acid (HOAC) and its conjugate base, acetate (OAc鈦�), we see a direct application of this concept. When mixed with the strong base NaOH, acetic acid donates a proton to form acetate, which is its conjugate base. The ability to identify conjugate acid-base pairs is crucial in buffer solutions as they are resistant to pH changes due to their ability to neutralize added acids or bases. This characteristic is woven into the fabric of the Henderson-Hasselbalch equation and is often exploited in buffer design and understanding the outcomes of various titrations.

Weak Acid & Strong Base Reaction
When a weak acid reacts with a strong base, a neutralization reaction occurs, forming water and the weak acid鈥檚 conjugate base. Unlike strong acids, weak acids do not dissociate completely in water; they establish an equilibrium that is shifted by the addition of a strong base. This characteristic leads to the formation of a buffer region where the pH changes gradually upon small additions of base or acid.

In our exercise example, acetic acid (HOAC), a weak acid, reacts with sodium hydroxide (NaOH), a strong base. The NaOH dissociates completely, yielding OH鈦� ions that react with HOAC, converting it to its conjugate base, acetate (OAc鈦�). The remaining unreacted HOAC and the newly formed OAc鈦� form a buffer solution that resists drastic pH changes, which is why its pH can be readily calculated using the Henderson-Hasselbalch equation. Such reactions are at the heart of acid-base chemistry and are pivotal in understanding the behavior of substances during titration and in buffer solutions.

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Most popular questions from this chapter

\(0.001 \mathrm{~mol}\) of \(\mathrm{NaOH}\) is added to \(100 \mathrm{ml}\). of a solution that is \(0.5 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) and \(0.5 \mathrm{M} \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\). Determine the \(\mathrm{pH}\) of this Solution; the \(\mathrm{K}_{\text {diss }}\) of acetic acid is \(1.8 \times 10^{-5}\).

For \(\mathrm{HF}, \mathrm{K}_{\text {diss }}=6.7 \times 10^{-4}\) what is the \(\mathrm{H}_{3} \mathrm{O}^{+}\) and \(\mathrm{OH}^{-}\) concentrations of a \(0.10 \mathrm{M} \mathrm{HF}\) solution? Assume \(\mathrm{K}_{\mathrm{W}}=1 \times 10^{-14}\)

Calculate the number of grams of \(\mathrm{NH}_{4} \mathrm{Cl}\) that must be added to \(2 \ell\) of \(0.1 \mathrm{M} \mathrm{NH}_{3}\) to prepare a solution of \(\mathrm{pH}=11.3\) at \(25^{\circ} \mathrm{C}\). The \(\mathrm{K}_{\mathrm{b}}\) for \(\mathrm{NH}_{3}=1.8 \times 10^{-5}\). The equilibrium equation is: \(\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \rightleftarrows \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-}\)

Assuming \(\mathrm{pD}=-\log \left[\mathrm{D}_{3} \mathrm{O}^{+}\right]\) in analogy to \(\mathrm{pH}\), what is \(\mathrm{pD}\) of pure \(\mathrm{D}_{2} \mathrm{O} ? \mathrm{~K}=2 \times 10^{-15}\).

Find the \(\mathrm{pH}\) of a \(0.2 \mathrm{M}\) solution of formic acid. \(\mathrm{K}_{\mathrm{a}}=1.76 \times 10^{-4}\)
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