Question

A buffer solution was prepared by adding \(4.95 \mathrm{~g}\) sodium acetate to \(250 . \mathrm{mL}\) of \(0.150-\mathrm{M}\) acetic acid. (a) What ions and molecules are present in the solution? List them in order of decreasing concentration. (b) Calculate the \(\mathrm{pH}\) of the buffer solution. (c) Calculate the pH of 100 . mL of the buffer solution if you add \(80 . \mathrm{mg} \mathrm{NaOH}\). (Assume negligible change in volume.) (d) Write a net ionic equation for the reaction that occurs to change the pH.

Short answer

(a) H2O, CH3COO-, CH3COOH, Na+, H+ (highest to lowest concentration). (b) pH = 4.95. (c) pH = 5.043 after adding NaOH. (d) CH3COOH + OH- → CH3COO- + H2O.

Step-by-step solution

Identify Ions and Molecules in the Buffer Solution

This buffer solution consists of acetic acid (\( ext{CH}_3 ext{COOH} \)) and sodium acetate (\( ext{CH}_3 ext{COONa} \)). In solution, sodium acetate dissociates to produce sodium ions (\( ext{Na}^+ \)) and acetate ions (\( ext{CH}_3 ext{COO}^- \)). Since it is a buffer solution, a significant amount of acetate ions from sodium acetate and acetic acid molecules will be present.The ions and molecules present in order of decreasing concentration are approximate as: 1. Water (\( ext{H}_2 ext{O} \))2. Acetate ions (\( ext{CH}_3 ext{COO}^- \))3. Acetic acid (\( ext{CH}_3 ext{COOH} \))4. Sodium ions (\( ext{Na}^+ \))5. Hydrogen ions (\( ext{H}^+ \)).

Calculate the Concentrations of Buffer Components

First, calculate the moles of sodium acetate:Molar mass of sodium acetate, \( ext{CH}_3 ext{COONa} = 82.03 ext{ g/mol} \).Moles of sodium acetate = \( \frac{4.95 ext{ g}}{82.03 ext{ g/mol}} = 0.0603 ext{ mol} \).Since the total volume is \( 0.250 ext{ L} \), the concentration of acetate ions from sodium acetate is \( \frac{0.0603}{0.250} = 0.2412 ext{ M} \).The concentration of acetic acid is given as \( 0.150 ext{ M} \).

Calculate the pH of the Buffer Solution Using the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is:\[ \text{pH} = \text{p}K_a + \log\left(\frac{[\text{Acetate ion}]}{[\text{Acetic Acid}]}\right) \]The \( K_a \) of acetic acid is \( 1.8 \times 10^{-5} \), thus \( ext{p}K_a = 4.74 \).Substitute values:\[ \text{pH} = 4.74 + \log\left(\frac{0.2412}{0.150}\right) = 4.74 + 0.209 = 4.95 \]

Calculate the pH Change After Adding NaOH

Convert the mass of \( ext{NaOH} \) added:\( ext{molar mass of NaOH} = 40.00 ext{ g/mol} \).Moles of \( ext{NaOH} \) = \( \frac{80 ext{ mg}}{40 ext{ g/mol}} = 0.002 ext{ mol} \).Adding \( ext{NaOH} \) decreases moles of \( ext{CH}_3 ext{COOH} \) and increases \( ext{CH}_3 ext{COO}^- \):New moles of acetic acid = \( 0.150 \times 0.100 = 0.015 - 0.002 = 0.013 ext{ mol} \).New moles of \( ext{CH}_3 ext{COO}^- \) = \( 0.2412 \times 0.100 + 0.002 = 0.02412 + 0.002 = 0.02612 ext{ mol} \).New concentrations in 100 mL solution:Acetic acid: \( 0.130 ext{ M} \)Acetate ion: \( 0.2612 ext{ M} \)Using the equation:\[ \text{pH} = 4.74 + \log\left(\frac{0.2612}{0.130}\right) = 4.74 + 0.303 = 5.043 \]

Write the Net Ionic Equation

The buffer reaction occurring when \( ext{NaOH} \) is added is:\[ \text{CH}_3 ext{COOH} + ext{OH}^- \rightarrow ext{CH}_3 ext{COO}^- + ext{H}_2 ext{O} \]This equation shows the conversion of acetic acid to acetate ions, neutralizing the added base.

Key concepts

Acid-Base Chemistry

Acid-base chemistry is fundamental in understanding buffer solutions. A buffer solution is a special type of solution that resists changes in pH when small amounts of an acid or a base are added. This ability is crucial in biological and chemical systems where stable pH levels are necessary. Buffer solutions typically consist of a weak acid and its conjugate base or a weak base and its conjugate acid.

For example, in the given exercise, acetic acid ( CH_3 COOH) acts as the weak acid, and sodium acetate ( CH_3 COONa) provides its conjugate base, acetate ion ( CH_3 COO^-). The interaction between these components helps maintain the pH of the solution when external substances are added. Understanding the basics of how acids and bases donate and accept protons will illuminate how buffer systems work.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a pivotal tool in acid-base chemistry, particularly in calculating the pH of buffer solutions. It is represented as:\[ \text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]where \( [\text{A}^-] \) is the concentration of the conjugate base, and \( [\text{HA}] \) is the concentration of the acid.

This equation allows you to relate the ratio of the concentrations of the deprotonated (base) and protonated (acid) forms of the buffer components to the solution's pH and the acid's pKa value. It simplifies the process of estimating the pH by using logarithmic functions, which handle the highly variable concentrations in a manageable linear form.

pH Calculation

Calculating the pH is a critical step in understanding a buffer solution's behavior. Initially, you use the Henderson-Hasselbalch equation with the concentrations you have calculated for both components of the buffer. For instance, in the exercise, calculate the initial pH using the concentrations derived:\[ \text{pH} = 4.74 + \log\left(\frac{0.2412}{0.150}\right) = 4.95 \]

When extra substances, like NaOH, are added, re-calculate using the new concentrations. This process reveals how the buffer withstands pH changes by neutralizing added acids or bases. Keeping a stable pH is essential, particularly in biological settings, as proteins and enzymes can be extremely pH-sensitive.

Ionic Reactions

Ionic reactions play a central role in buffer systems. Consider the net ionic equation:\[ \text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \]This equation exemplifies how a buffer system responds when a strong base like NaOH is introduced.

The buffer reacts by converting the added hydroxide ions (\(\text{OH}^-\)) with acetic acid, forming acetate ions and water. This mechanism prevents a drastic increase in pH. Such reactions showcase the buffer's capacity to 'absorb' the external ions' impact, maintaining equilibrium and ensuring that changes in pH remain minimal. Comprehending this reaction aids in recognizing the buffering action as a dynamic chemical balance rather than a static process.