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\(K_{a}\) for acetic acid is \(1.7 \times 10^{-5}\) at \(25^{\circ} \mathrm{C}\). A buffe solution is made by mixing \(52.1 \mathrm{~mL}\) of \(0.122 \mathrm{M}\) acetic acic with \(46.1 \mathrm{~mL}\) of \(0.182 \mathrm{M}\) sodium acetate. Calculate the \(\mathrm{pH}\) of this solution at \(25^{\circ} \mathrm{C}\) after the addition of \(5.82 \mathrm{~mL}\) of \(0.125 \mathrm{M} \mathrm{NaOH}\)

Short Answer

Expert verified
The pH of the buffer solution is 4.98.

Step by step solution

01

Calculate Initial Moles of Acetic Acid and Sodium Acetate

First, calculate the initial moles of acetic acid (CH鈧僀OOH) and sodium acetate (CH鈧僀OONa) using the given volumes and concentrations.For acetic acid:\[\text{Moles of CH}_3\text{COOH} = 0.122 \text{ M} \times 0.0521 \text{ L} = 0.0063562 \text{ mol}\]For sodium acetate:\[\text{Moles of CH}_3\text{COONa} = 0.182 \text{ M} \times 0.0461 \text{ L} = 0.0083902 \text{ mol}\]
02

Calculate Moles of NaOH Added

Determine the moles of NaOH added to the solution.\[\text{Moles of NaOH} = 0.125 \text{ M} \times 0.00582 \text{ L} = 0.0007275 \text{ mol}\]
03

Determine the New Concentrations After NaOH Addition

Since NaOH will react with acetic acid to form more acetate, calculate the remaining moles of acetic acid and new amount of acetate formed.For acetic acid after reaction:\[\text{Moles of CH}_3\text{COOH remaining} = 0.0063562 - 0.0007275 = 0.0056287 \text{ mol}\]For acetate after reaction:\[\text{Moles of CH}_3\text{COO}^- = 0.0083902 + 0.0007275 = 0.0091177 \text{ mol}\]
04

Calculate Total Volume of the Solution

Compute the total volume of the buffer solution after addition of NaOH.\[\text{Total Volume} = 52.1 \text{ mL} + 46.1 \text{ mL} + 5.82 \text{ mL} = 104.02 \text{ mL} = 0.10402 \text{ L}\]
05

Calculate Equilibrium Concentrations

Determine the equilibrium concentrations of the acetic acid and acetate ion.For acetic acid:\[[\text{CH}_3\text{COOH}] = \frac{0.0056287}{0.10402} = 0.0541 \text{ M}\]For acetate ion:\[[\text{CH}_3\text{COO}^-] = \frac{0.0091177}{0.10402} = 0.0876 \text{ M}\]
06

Apply the Henderson-Hasselbalch Equation

Use the Henderson-Hasselbalch equation to find the pH of the buffer solution:\[\text{pH} = \text{p}K_a + \log\left(\frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}\right)\]First, calculate \(\text{p}K_a\):\[\text{p}K_a = -\log(1.7 \times 10^{-5}) = 4.77\]Then, calculate the pH:\[\text{pH} = 4.77 + \log\left(\frac{0.0876}{0.0541}\right) = 4.77 + 0.210 = 4.98\]

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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 handy formula used mainly in biochemistry and chemistry to calculate the pH of buffer solutions. A buffer solution contains a weak acid and its conjugate base. This equation relates the acidity (pH) to the concentration of the acid ( ext{[HA]}), its conjugate base ( ext{[A^-]}), and the acid's ionization constant (pK鈧). The equation is expressed as follows: \[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]This comes in handy when you know the quantities of an acid and its conjugate base in a solution because it directly links these concentrations to the pH - a crucial indicator of acidity or alkalinity.
This equation assumes that the concentrations of acidic and basic components do not change significantly upon decomposition, making it perfect for buffer solutions as they resist changes in pH when small amounts of acid or base are added.
Acetic Acid
Acetic acid, also known chemically as CH鈧僀OOH, is a typical example of a weak acid. It's a colorless, pungent liquid that is present in vinegar at about 4-7% concentration. - **Weak Acid**: Acetic acid partially ionizes in a solution, meaning only some percentage of its molecules donate H鈦 ions to the solution.- **Acid Dissociation Constant \( K_a \)**: The acid dissociation constant \(K_a\) for acetic acid is a measure of its degree of ionization, which is \(1.7 \times 10^{-5}\). This relatively low \(K_a\) indicates that it is not fully energetic at ionizing, hence the classification as a weak acid.
In a buffer solution comprising acetic acid and its conjugate base, like the one described in this exercise, acetic acid plays a critical role in maintaining a stable pH by accepting H鈦 from added bases.
Sodium Acetate
Sodium acetate ( ext{CH}_3 ext{COONa}) is the sodium salt of acetic acid, featuring prominently as the conjugate base. It is often found in this form: - **Conjugate Base**: In a solution, sodium acetate dissociates into sodium ions ( ext{Na}^+) and acetate ions ( ext{CH}_3 ext{COO}^-). The acetate ions are the key players here, acting as the conjugate base in our buffer solution. - **Buffer Action**: They act to neutralize added acids, combining with H鈦 ions to reform acetic acid, thus preventing any significant change in the pH.
The equilibrium between sodium acetate and acetic acid in a solution is crucial for sustaining pH levels. It provides a resistance to drastic changes by alternately accepting or donating H鈦 ions, hence facilitating the buffering action.
pH Calculation
The process of calculating the pH of a buffer solution involves several steps but relies heavily on the use of the Henderson-Hasselbalch equation. Let's break it down:1. **Calculate Initial Concentrations**: First, determine the concentration of the weak acid (acetic acid) and the conjugate base (sodium acetate) after mixing.2. **Account for Reactions**: Consider how the addition of other substances like ext{NaOH} could affect these concentrations, as NaOH reacts with acetic acid, altering initial amounts.3. **Use the Equation**: Plug these equilibrium concentrations into the Henderson-Hasselbalch equation, alongside the ext{pK鈧恾 of acetic acid (which is calculated as \ \log(1.7 \times 10^{-5}) = 4.77).
Working through these steps for the example provided, the pH can be determined as follows:\[ \text{pH} = 4.77 + \log\left(\frac{0.0876}{0.0541}\right) = 4.98 \]This means the buffer resists significant pH changes, remaining around 4.98 even when NaOH is added.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of a solution made up from \(2.0 \mathrm{~g}\) of potassium hydroxide dissolved in \(115 \mathrm{~mL}\) of \(0.19 \mathrm{M}\) perchloric acid. Assume the change in volume due to adding potassium hydroxide is negligible.

A generic base, \(\mathrm{B}^{-},\) is added to \(2.25 \mathrm{~L}\) of water. The \(\mathrm{pH}\) of the solution is found to be 10.10 . What is the concentration of the base \(\mathrm{B}^{-}\) in this solution? \(K_{a}\) for the acid \(\mathrm{HB}\) at \(25^{\circ} \mathrm{C}\) is \(1.99 \times 10^{-9}\)

Rantidine is a nitrogen base that is used to control stomach acidity by suppressing the stomach's production of hydrochloric acid. The compound is present in Zantac \(^{\mathbb{B}}\) as the chloride salt (rantidinium chloride; also called rantidine hydrochloride). Do you expect a solution of rantidine hydrochloride to be acidic, basic, or neutral? Explain by means of a general chemical equation.

Malic acid is a weak diprotic organic acid with \(K_{a 1}=4.0 \times 10^{-4}\) and \(K_{a 2}=9.0 \times 10^{-6}\) a. Letting the symbol \(\mathrm{H}_{2}\) A represent malic acid, write the chemical equations that represent \(K_{a 1}\) and \(K_{a 2}\). Write the chemical equation that represents \(K_{a 1} \times K_{a 2}\) b. Qualitatively describe the relative concentrations of \(\mathrm{H}_{2} \mathrm{~A}, \mathrm{HA}^{-}, \mathrm{A}^{2-},\) and \(\mathrm{H}_{3} \mathrm{O}^{+}\) in a solution that is about one molar in malic acid. c. Calculate the \(\mathrm{pH}\) of a \(0.0175 \mathrm{M}\) malic acid solution and the equilibrium concentration of \(\left[\mathrm{H}_{2} \mathrm{~A}\right]\) d. What is the \(\mathrm{A}^{2-}\) concentrationin in solutions \(\mathrm{D}\) and \(\mathrm{c}\) ?

Find the \(\mathrm{pH}\) of the solution obtained when \(25 \mathrm{~mL}\) of \(0.065 \mathrm{M}\) benzylamine, \(\mathrm{C}_{7} \mathrm{H}_{7} \mathrm{NH}_{2},\) is titrated to the equivalence point with \(0.050 M\) hydrochloric acid. \(K_{b}\) for benzylamine is \(4.7 \times 10^{-10}\)

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