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Chapter 16: Problem 79

What is the \(\mathrm{pH}\) of a buffer solution that is \(0.15 \mathrm{M}\) chloroacetic acid and \(0.10 M\) sodium chloroacetate? \(K_{a}=\) \(1.3 \times 10^{-3}\)

Short Answer

Expert verified
The pH of the buffer solution is approximately 2.71.

Step by step solution

01

Understanding the Problem

We need to find the pH of a buffer solution containing chloroacetic acid and sodium chloroacetate. This can be done using the Henderson-Hasselbalch equation, involving the acid dissociation constant \(K_a\) and the concentrations of the acid and its conjugate base.
02

Writing the Henderson-Hasselbalch Equation

The \[\text{Henderson-Hasselbalch equation} \text{ is } \]\[\mathrm{pH} = \mathrm{p}K_a + \log \left(\frac{[\text{Base}]}{[\text{Acid}]\right)\]Where \[\mathrm{p}K_a = -\log{K_a}\] and \([\text{Base}]\) is the concentration of the sodium chloroacetate, \([\text{Acid}]\) is the concentration of the chloroacetic acid.
03

Calculating the pKa

Calculate \(\mathrm{p}K_a\) using the formula \[\mathrm{p}K_a = -\log{(K_a)}\]. Thus, \(\mathrm{p}K_a = -\log{(1.3 \times 10^{-3})}\approx 2.886\).
04

Calculating the pH of the Buffer Solution

Use the Henderson-Hasselbalch equation to calculate the pH:\[\mathrm{pH} = 2.886 + \log \left(\frac{0.10}{0.15}\right)\]\(\Rightarrow \mathrm{pH} = 2.886 + \log(0.6667)\) which is approximately \(\mathrm{pH} = 2.886 - 0.176\).Therefore, \(\mathrm{pH} \approx 2.710\).

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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 tool used in chemistry to estimate the pH of a buffer solution. This equation makes it easy to calculate the pH based on the known concentrations of an acid and its conjugate base, combined with the acid's known dissociation constant, noted as \( K_a \). The equation is expressed as follows:\[\mathrm{pH} = \mathrm{p}K_a + \log \left(\frac{[\text{Base}]}{[\text{Acid}]}\right)\]This equation is particularly useful because it simplifies the process of finding pH without needing to solve quadratic equations. **Key Components** include:
  • \( \mathrm{p}K_a \): The negative logarithm of the acid dissociation constant \( K_a \), calculated using \( \mathrm{p}K_a = -\log(K_a) \).
  • \( [\text{Base}] \): Concentration of the base form (sodium chloroacetate in this case).
  • \( [\text{Acid}] \): Concentration of the acidic form (chloroacetic acid).
These elements allow chemists to determine the pH of solutions where both acid and base coexist, known as buffer solutions. Remember, buffers resist drastic pH changes, which is vital in many chemical and biological systems.
Acid-Base Chemistry
In the realm of chemistry, acid-base chemistry is crucial to understand how substances interact in solutions. **Acids and bases** are defined by their ability to donate or accept protons, based on the Br酶nsted-Lowry theory.- **Acids** are substances that donate protons (\( \text{H}^+ \)).- **Bases** accept protons, generating their conjugate acids.A buffer is a solution that contains a weak acid and its conjugate base. It can maintain a constant pH, even with additions of small amounts of acid or base. This characteristic is essential for many biological processes, like human blood which is buffered to maintain pH levels.In buffer solutions:
  • The acidic component reacts with added bases to resist pH changes.
  • The base component reacts with added acids.
This equilibrium between an acid and its conjugate base allows the solution to neutralize small disturbances in pH, thus performing a shielding role in chemical reactions and biological systems.
Chloroacetic Acid
Chloroacetic acid is a substituted carboxylic acid, characterized by one chlorine atom attached to its acetic acid's methyl group. It is noted for being a stronger acid than acetic acid due to the electron-withdrawing chlorine, which stabilizes its conjugate base.**Properties of Chloroacetic Acid:**
  • **Chemical Formula:** \( \text{C}_2\text{H}_3\text{ClO}_2 \)
  • **Acidity:** Due to the chlorine atom, it has a higher acidity, with a lower \( pK_a \) than acetic acid.
  • **Usage:** Often used in the synthesis of various chemical compounds, including dyes and pharmaceuticals.
Chloroacetic acid is crucial in buffer solutions, such as in the given exercise, to provide a stable pH environment. This particular acid exemplifies the concept of how structural changes in a molecule can affect its acidity, demonstrating how electron-withdrawing groups influence acid strength. Such understanding is fundamental in designing effective buffers for industrial or research purposes.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of a solution made by mixing \(0.60 \mathrm{~L}\) of \(0.10 M \mathrm{NH}_{4} \mathrm{Cl}\) with \(0.50 \mathrm{~L}\) of \(0.10 \mathrm{M} \mathrm{NaOH} . K_{b}\) for \(\mathrm{NH}_{3}\) is \(1.8 \times 10^{-5}\)

Tartaric acid is a weak diprotic fruit acid with \(K_{a 1}=\) \(1.0 \times 10^{-3}\) and \(K_{a 2}=4.6 \times 10^{-5}\) a. Letting the symbol \(\mathrm{H}_{2} \mathrm{~A}\) represent tartaric 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 \(0.5 \mathrm{M}\) in tartaric acid. c. Calculate the \(\mathrm{pH}\) of a \(0.0250 \mathrm{M}\) tartaric acid solution and the equilibrium concentration of \(\left[\mathrm{H}_{2} \mathrm{~A}\right]\) d. What is the \(A^{2-}\) concentration?

\(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NH}_{2} \mathrm{COOH}\), para- aminobenzoic acid \((\mathrm{PABA})\), is used in some sunscreen agents. Calculate the concentrations of hydronium ion and para-aminobenzoate ion, \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NH}_{2} \mathrm{COO}^{-}\), in a \(0.055 M\) solution of the acid. The value of \(K_{a}\) is \(2.2 \times 10^{-5}\).

Consider a solution of \(0.0010 M\) HF \(\left(K_{a}=6.8 \times 10^{-4}\right)\). In solving for the concentrations of species in this solution, could you use the simplifying assumption in which you neglect \(x\) in the denominator of the equilibrium equation? Explain.

The equilibrium equations and \(K_{a}\) values for three reaction systems are given below. \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}\) $$ \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{HC}_{2} \mathrm{O}_{4}^{-}(a q) ; K_{a}=5.6 \times 10^{-2} $$ \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}=\) $$ \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) ; K_{a}=6.9 \times 10^{-3} $$ \(\mathrm{HCOOH}(a q)+\mathrm{H}_{2} \mathrm{O}=\) $$ \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{HCOO}^{-}(a q) ; K_{a}=1.7 \times 10^{-4} $$ a. Which conjugate pair would be best for preparing a buffer with a pH of \(2.88\) ? b. How would you prepare \(50 \mathrm{~mL}\) of a buffer with a pH of \(2.88\) assuming that you had available \(0.10 \mathrm{M}\) solutions of each pair?
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