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Chapter 17: Problem 23

A buffer is prepared by adding \(20.0 \mathrm{~g}\) of sodium acetate \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\) to \(500 \mathrm{~mL}\) of a \(0.150 \mathrm{M}\) acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) solution. (a) Determine the \(\mathrm{pH}\) of the buffer. (b) Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of sodium hydroxide solution are added to the buffer.

Short Answer

Expert verified
The pH of the buffer solution is approximately 5.76. The complete ionic equation for the reaction between hydrochloric acid and buffer components is \(H^{+}(aq) + CH_3COO^{-}(aq) \rightarrow CH_3COOH(aq)\), and for the reaction between sodium hydroxide and buffer components is \(OH^{-}(aq) + CH_3COOH(aq) \rightarrow CH_3COO^{-}(aq) + H_2O(l)\).

Step by step solution

01

1. Calculate the moles of sodium acetate and acetic acid in the buffer solution

To calculate the moles of sodium acetate, we can use the given mass and molecular weight of the substance: Moles of sodium acetate = (20.0 g) / (molecular weight of CH鈧僀OONa) Molecular weight of CH鈧僀OONa = (12.01 * 2) + (1.01 * 3) + (16.00 * 2) + (22.99 * 1) = 82.04 g/mol Moles of sodium acetate = 20.0 g / 82.04 g/mol 鈮� 0.244 mol Now, let's calculate the moles of acetic acid: Moles of acetic acid = (0.150 mol/L) x (0.500 L) = 0.075 mol
02

2. Use the Henderson-Hasselbalch equation to determine the pH of the buffer

The Henderson-Hasselbalch equation is: pH = pKa + log ([A鈦籡 / [HA]) We're given that the acetic acid (CH鈧僀OOH) has a pKa = 4.75. We have moles of sodium acetate (A鈦�) and moles of acetic acid (HA), so we can calculate the pH: pH = 4.75 + log (0.244 mol / 0.075 mol) 鈮� 4.75 + 1.01 鈮� 5.76 Thus, the pH of the buffer solution is approximately 5.76.
03

3. Write the complete ionic equation for the reaction between hydrochloric acid and the buffer constituents

When hydrochloric acid (HCl) is added to the buffer, it reacts with the acetate ion (CH鈧僀OO鈦�) from sodium acetate to form acetic acid: H鈦�(aq) + Cl鈦�(aq) + CH鈧僀OO鈦�(aq) -> CH鈧僀OOH(aq) + Cl鈦�(aq) The complete ionic equation is: H鈦�(aq) + CH鈧僀OO鈦�(aq) -> CH鈧僀OOH(aq)
04

4. Write the complete ionic equation for the effect that occurs when sodium hydroxide is added to the buffer

When sodium hydroxide (NaOH) is added to the buffer, it reacts with the undissociated acetic acid (CH鈧僀OOH) to form sodium acetate: Na鈦�(aq) + OH鈦�(aq) + CH鈧僀OOH(aq) -> CH鈧僀OO鈦�(aq) + H鈧侽(l) + Na鈦�(aq) The complete ionic equation is: OH鈦�(aq) + CH鈧僀OOH(aq) -> CH鈧僀OO鈦�(aq) + H鈧侽(l) To summarize, the pH of the buffer solution is approximately 5.76. The complete ionic equation for the reaction between hydrochloric acid and buffer components is H鈦�(aq) + CH鈧僀OO鈦�(aq) -> CH鈧僀OOH(aq), and for the reaction between sodium hydroxide and buffer components is OH鈦�(aq) + CH鈧僀OOH(aq) -> CH鈧僀OO鈦�(aq) + H鈧侽(l).

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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 fundamental in understanding how buffer solutions work. It is a straightforward formula that relates the pH of a buffer solution to the concentration of acid and its conjugate base.
The equation is given by:\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]Here, \(\text{pK}_a\) is the negative logarithm (base 10) of the acid dissociation constant (\(K_a\)) for the acid, \([\text{HA}]\) represents the concentration of the undissociated acid, and \([\text{A}^-]\) represents the concentration of the conjugate base.

A buffer solution consists of a weak acid and its conjugate base. In our example, the weak acid is acetic acid (CH鈧僀OOH), and the conjugate base is acetate (CH鈧僀OO鈦�).
The Henderson-Hasselbalch equation helps in calculating the pH of the buffer by considering the relative concentrations of these two components.
Acetic Acid
Acetic acid is a common choice when creating buffer solutions due to its weak acidic properties.
It's composed of the chemical formula CH鈧僀OOH. Among its unique characteristics, acetic acid partially dissociates in water, making it suitable as a component of buffer solutions.

In our problem scenario, acetic acid provides the proton (H鈦�) in the buffer solution, interacting with acetate ions provided by sodium acetate. This interaction is crucial in maintaining the pH of the solution.
When you dissolve acetic acid in water, it establishes an equilibrium between its undissociated form (CH鈧僀OOH) and the dissociated ions (CH鈧僀OO鈦� and H鈦�). This equilibrium is critical in its ability to resist changes in pH when acids or bases are added to the solution. The balance between these forms is what grants the buffer its effectiveness.
Ionic Equations
Writing complete ionic equations is an important step in understanding buffer reactions. These equations illustrate the species present in a reaction at the ionic level.

When hydrochloric acid (HCl) is added to a buffer solution containing acetic acid, the hydrogen ion (H鈦�) from the HCl reacts with the acetate ion (CH鈧僀OO鈦�) to form more acetic acid.
The complete ionic equation simplifies this as:\[ \text{H}^+(\text{aq}) + \text{CH}_3\text{COO}^-(\text{aq}) \rightarrow \text{CH}_3\text{COOH}(\text{aq}) \]Similarly, when sodium hydroxide (NaOH) is added, the hydroxide ion (OH鈦�) reacts with acetic acid to generate more acetate ions and water:\[ \text{OH}^-(\text{aq}) + \text{CH}_3\text{COOH}(\text{aq}) \rightarrow \text{CH}_3\text{COO}^-(\text{aq}) + \text{H}_2\text{O}(\text{l}) \]Understanding these ionic interactions is crucial in grasping how buffers can neutralize added acids and bases, thus maintaining a stable pH in solution.

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Most popular questions from this chapter

The solubility-product constant for barium permanganate, \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\), is \(2.5 \times 10^{-10}\). Assume that solid \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\) is in equilibrium with a solution of \(\mathrm{KMnO}_{4}\). What concentration of \(\mathrm{KMnO}_{4}\) is required to establish a concentration of \(2.0 \times 10^{-8} \mathrm{M}\) for the \(\mathrm{Ba}^{2+}\) ion in solution?

Consider a beaker containing a saturated solution of \(\mathrm{CaF}_{2}\) in equilibrium with undissolved \(\mathrm{CaF}_{2}(s) .\) (a) If solid \(\mathrm{CaCl}_{2}\) is added to this solution, will the amount of solid \(\mathrm{CaF}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Ca}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of \(\mathrm{F}^{-}\) ions in solution increase or decrease?

Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\) and whose corresponding \(K_{s p}=6.8 \times 10^{-27} .\) As discussed in the "Chemistry and Life" box on page 730 , fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F},\) whose \(K_{s p}=1.0 \times 10^{-60} \cdot(\mathrm{a})\) Write the expres- sion for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

(a) Calculate the pH of a buffer that is \(0.12 \mathrm{M}\) in lactic acid and \(0.11 M\) in sodium lactate. (b) Calculate the pH of a buffer formed by mixing \(85 \mathrm{~mL}\) of \(0.13 \mathrm{M}\) lactic acid with \(95 \mathrm{~mL}\) of \(0.15 \mathrm{M}\) sodium lactate.

Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONa) to enough water to make \(1.00 \mathrm{~L}\) of solution. Buffer \(\mathrm{A}\) is prepared using \(1.00 \mathrm{~mol}\) each of formic acid and sodium formate. Buffer B is prepared by using \(0.010 \mathrm{~mol}\) of each. (a) Calculate the \(\mathrm{pH}\) of each buffer, and explain why they are equal. (b) Which buffer will have the greater buffer capacity? Explain. (c) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(1.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). (d) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(10 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). (e) Discuss your answers for parts (c) and (d) in light of your response to part (b).
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