Chapter 16: Problem 63
What is the original molarity of a solution of formic acid \((\mathrm{HCOOH})\) whose \(\mathrm{pH}\) is 3.26 at \(25^{\circ} \mathrm{C} ?\left(K_{\mathrm{a}}\right.\) for $$ \text { formic acid } \left.=1.7 \times 10^{-4} .\right) $$
Short Answer
Expert verified
The original molarity of the solution is approximately 1.78 mM.
Step by step solution
01
Understand the Formic Acid Dissociation
Formic acid \(\mathrm{HCOOH}\) dissociates in water according to the equilibrium reaction: \(\mathrm{HCOOH}(aq) \rightleftharpoons \mathrm{H}^+(aq) + \mathrm{HCOO}^-(aq)\).The acid dissociation constant \(K_a\) is given by: \[ K_a = \frac{[\mathrm{H}^+][\mathrm{HCOO}^-]}{[\mathrm{HCOOH}]} = 1.7 \times 10^{-4} \]. \
02
Calculate the Concentration of \( [\mathrm{H}^+] \)
The \(\mathrm{pH}\) of the solution is 3.26. We can find the concentration of hydrogen ions \(\mathrm{H}^+\) using the \(\mathrm{pH}\) formula:\[ \mathrm{pH} = -\log [\mathrm{H}^+] \].Plugging in the \(\mathrm{pH}\) value, we have:\[ 3.26 = -\log [\mathrm{H}^+] \].Converting the \(\mathrm{pH}\) back to concentration:\[ [\mathrm{H}^+] = 10^{-3.26} \approx 5.50 \times 10^{-4} \text{ M} \]. \
03
Set Up the Expression for \( K_a \)
Using the expression for \(K_a\),\[ K_a = \frac{[\mathrm{H}^+][\mathrm{HCOO}^-]}{[\mathrm{HCOOH}]} = 1.7 \times 10^{-4} \].Assuming that \(x\) is the concentration of \(\mathrm{H}^+\) and \(\mathrm{HCOO}^-\) at equilibrium, we plug \(x = 5.50 \times 10^{-4}\) into the expression:\[ 1.7 \times 10^{-4} = \frac{(5.50 \times 10^{-4})^2}{C - 5.50 \times 10^{-4}} \]where \(C\) is the original molarity of formic acid. \
04
Solve for Initial Concentration \( C \)
Now, solve for \(C\):Assume \(C - 5.50 \times 10^{-4} \approx C\) because \(5.50 \times 10^{-4}\) is small compared to \(C\). This is a valid assumption when the ion concentration is much smaller than the initial concentration.Plugging values back, we simplify to:\[ 1.7 \times 10^{-4} = \frac{(5.50 \times 10^{-4})^2}{C} \].Multiply both sides by \(C\) and rearrange to solve for \(C\):\[ C = \frac{(5.50 \times 10^{-4})^2}{1.7 \times 10^{-4}} \approx 1.78 \times 10^{-3} \text{ M} \]. \
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Acid Dissociation Constant
Understanding the acid dissociation constant, often represented as \( K_a \), is crucial for analyzing how acids behave in solution. It's a measure of the strength of an acid, specifically how well an acid dissociates into its components—hydrogen ions (\( H^+ \)) and its conjugate base (\( ext{HCOO}^- \) for formic acid) in water. When formic acid (\( ext{HCOOH} \)) is dissolved:
- It partly dissociates to form \( H^+ \) and \( ext{HCOO}^- \).
- The equilibrium expression for this dissociation is:
pH Calculation
The \( ext{pH} \) scale serves as a quick reference to the acidity or basicity of a solution. It's derived from the concentration of hydrogen ions \( \text{H}^+ \) in the solution. To calculate \( \text{pH} \), the formula is:
- \( \text{pH} = -\log [\text{H}^+] \)
- Solve \( 3.26 = -\log [\text{H}^+] \).
- This leads to \( [\text{H}^+] = 10^{-3.26} \).
- Which simplifies to approximately \( 5.50 \times 10^{-4} \text{ M} \).
Equilibrium Reaction
In the context of acid dissociation, equilibrium reactions describe the balance between the reactants and products as the reaction reaches a state of dynamic equilibrium. This means the forward reaction rate (acid dissociating) equals the reverse rate (re-association of ions). For formic acid:
- The dissociation reaction is \( \text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^- \).
- Initially, as the acid starts to dissociate, \( \text{H}^+ \) and \( \text{HCOO}^- \) concentrations increase.
- Equilibrium is reached when their concentrations stabilize.