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Chloroacetic acid, \(\mathrm{ClCH}_{2} \mathrm{CO}_{2} \mathrm{H}\), is a moderately weak acid \(\left(K_{\mathrm{a}}=1.40 \times 10^{-3}\right) .\) If you dissolve \(94.5 \mathrm{mg}\) of the acid in water to give 125 mL of solution, what is the \(\mathrm{pH}\) of the solution?

Short Answer

Expert verified
The pH of the solution is approximately 2.47.

Step by step solution

01

Calculate the Molar Mass of Chloroacetic Acid

First, we need to determine the molar mass of chloroacetic acid (\(\mathrm{ClCH}_{2}\mathrm{CO}_{2}\mathrm{H}\)). The molar mass is calculated by adding the atomic masses of each element present in the compound: \(\text{Cl} (35.45) + 2 \times \text{H} (1.01) + 2 \times \text{O} (16.00) + \text{C} (12.01) = 94.49 \, \text{g/mol}\).
02

Calculate the Moles of Chloroacetic Acid

Convert the mass of chloroacetic acid from milligrams to grams: \(94.5 \, \mathrm{mg} = 0.0945 \, \mathrm{g}\). Next, calculate the number of moles by dividing the mass by the molar mass that was calculated in Step 1: \[\text{Moles} = \frac{0.0945 \, \text{g}}{94.49 \, \text{g/mol}} = 0.001001 \, \text{mol}.\]
03

Calculate the Concentration of the Acid

The concentration of chloroacetic acid is found by dividing the number of moles by the volume of the solution in liters. Convert the volume from mL to L: \(125 \, \mathrm{mL} = 0.125 \, \mathrm{L}\). Then, \[\text{Concentration} = \frac{0.001001 \, \text{mol}}{0.125 \, \text{L}} = 0.008008 \, \text{M}.\]
04

Set Up the Equilibrium Expression for Acid Dissociation

Chloroacetic acid dissociates in water according to: \(\mathrm{ClCH}_{2}\mathrm{CO}_{2}\mathrm{H} \rightleftharpoons \mathrm{ClCH}_{2}\mathrm{CO}_{2}^{-} + \mathrm{H}^{+}\). The expression for \(K_a\) is: \(K_a = \frac{[\mathrm{ClCH}_2\mathrm{CO}_2^{-}][\mathrm{H}^+]}{[\mathrm{ClCH}_2\mathrm{CO}_2\mathrm{H}]}\).
05

Assume Initial Concentrations for Equilibrium Calculations

Assume initial concentration of \([\mathrm{H}^{+}] = x\). Then for \([\mathrm{ClCH}_{2}\mathrm{CO}_{2}^{-}] = x\) and \([\mathrm{ClCH}_2\mathrm{CO}_2\mathrm{H}] = 0.008008 - x\). Since \(K_a\) is small, assume \(x\) is small compared to the initial concentration. Thus, \([\mathrm{ClCH}_2\mathrm{CO}_2\mathrm{H}] \approx 0.008008\).
06

Solve for \([\mathrm{H}^+]\) Concentration

Substitute the concentrations into the \(K_a\) expression: \[1.40 \times 10^{-3} = \frac{x^2}{0.008008}\]. Solve for \(x\): \[x^2 = (1.40 \times 10^{-3}) \times 0.008008 \approx 1.12112 \times 10^{-5}\]. \[x = \sqrt{1.12112 \times 10^{-5}} \approx 0.00335\]. Thus, \([\mathrm{H}^+] = 0.00335 \, \text{M}\).
07

Calculate the pH of the Solution

The \(\mathrm{pH}\) is calculated using the formula \(\mathrm{pH} = -\log_{10}[\mathrm{H}^+]\). Substituting the \([\mathrm{H}^+] = 0.00335 \, \text{M}\), \[\mathrm{pH} = -\log_{10}(0.00335) \approx 2.47\].

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Key Concepts

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

Molar Mass Calculation
Calculating the molar mass of a compound is the first step in many chemistry problems. Here, we focus on chloroacetic acid, represented as \(\mathrm{ClCH}_{2}\mathrm{CO}_{2}\mathrm{H}\). To find its molar mass, we sum the atomic masses of each constituent atom:
  • Chlorine (Cl): 35.45 g/mol
  • Hydrogen (H): 2 atoms at 1.01 g/mol each
  • Oxygen (O): 2 atoms at 16.00 g/mol each
  • Carbon (C): 12.01 g/mol
Adding these together gives us a total molar mass of 94.49 g/mol.
This value allows us to convert between grams and moles, a critical step in many chemical equilibrium and concentration calculations.
Equilibrium Expression
In chemistry, when a weak acid like chloroacetic acid dissolves in water, it only partially dissociates. This dissociation is represented by the equilibrium expression:
\[\mathrm{ClCH}_{2}\mathrm{CO}_{2}\mathrm{H} \rightleftharpoons \mathrm{ClCH}_{2}\mathrm{CO}_{2}^{-} + \mathrm{H}^{+}\]
To describe this equilibrium mathematically, we use the acid dissociation constant, \(K_a\), given by:
\[K_a = \frac{[\mathrm{ClCH}_2\mathrm{CO}_2^{-}][\mathrm{H}^+]}{[\mathrm{ClCH}_2\mathrm{CO}_2\mathrm{H}]}\]
This equation shows the balance between the dissociated ions and the undissociated acid. It's essential to understand how these concentrations relate to reactants and products in weak acid reactions.
The small \(K_a\) value indicates that the acid is indeed weak, meaning it does not completely ionize in solution.
pH Calculation
pH is a measure of the acidity or basicity of a solution. For our weak acid, after dissolution and establishing equilibrium, we calculate \([\mathrm{H}^{+}]\) concentration.
This allows us to compute the solution's pH using the formula:
\[\mathrm{pH} = -\log_{10}[\mathrm{H}^+]\]
By substituting the calculated \([\mathrm{H}^+]\) value of 0.00335 M, we find:
\[\mathrm{pH} = -\log_{10}(0.00335) \approx 2.47\]
This value tells us that the solution is acidic, as expected from the dissociation of chloroacetic acid, even though it is considered a moderately weak acid.
Weak Acid Dissociation Constant
The dissociation constant \(K_a\) is significant in understanding weak acids like chloroacetic acid. It quantifies the degree of ionization and provides insights into the acid's strength.
The given \(K_a\) for chloroacetic acid is \(1.40 \times 10^{-3}\), indicating partial dissociation in water. This small value shows that not all of the acid molecules give away their protons (\(\mathrm{H}^+\)) to form the ions
\([\mathrm{ClCH}_2\mathrm{CO}_2^{-}]\) and \([\mathrm{H}^+]\).
  • A larger \(K_a\) would mean a stronger acid (more dissociation).
  • A smaller \(K_a\) suggests a weaker acid (less dissociation).
Understanding \(K_a\) helps predict how much the weak acid will affect the pH of a solution.

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Most popular questions from this chapter

Explain why benzenesulfonic acid is a Bronsted acid.

Several acids are listed here with their respective equilibrium constants. $$\begin{aligned} &\mathrm{HF}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{F}^{-}(\mathrm{aq})\\\ &K_{\mathrm{a}}=7.2 \times 10^{-4} \end{aligned}$$ $$\begin{aligned} &\mathrm{HPO}_{4}^{2-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{PO}_{4}^{3-}(\mathrm{aq})\\\ &K_{\mathrm{a}}=3.6 \times 10^{-13} \end{aligned}$$ $$\begin{aligned} &\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(\mathrm{aq})\\\ &K_{\mathrm{a}}=1.8 \times 10^{-5} \end{aligned}$$ (a) Which is the strongest acid? Which is the weakest acid? (b) What is the conjugate base of the acid HF? (c) Which acid has the weakest conjugate base? (d) Which acid has the strongest conjugate base?

What is the \(\mathrm{pH}\) of a \(0.0075 \mathrm{M}\) solution of HCl? What is the hydroxide ion concentration of the solution?

A \(0.10 \mathrm{M}\) solution of chloroacetic acid, \(\mathrm{ClCH}_{2} \mathrm{CO}_{2} \mathrm{H},\) has a pH of \(1.95 .\) Calculate \(K_{\mathrm{a}}\) for the acid.

Which is the stronger of the following two acids? (a) acetic acid, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=1.8 \times 10^{-5}\) (b) chloroacetic acid, \(\mathrm{ClCH}_{2} \mathrm{CO}_{2} \mathrm{H}, \mathrm{p} K_{\mathrm{a}}=2.85\)

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