Problem 63 - Chemistry

Question

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 $$ \text { formic acid } \left.=1.7 \times 10^{-4} .\right) $$

Step-by-Step Solution

Answer

The original molarity of the solution is approximately 1.78 mM.

Step 1: 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: \[ K_a = \frac{[\mathrm{H}^+][\mathrm{HCOO}^-]}{[\mathrm{HCOOH}]} = 1.7 \times 10^{-4} \]. \

Step 2: 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:\[ [\mathrm{H}^+] = 10^{-3.26} \approx 5.50 \times 10^{-4} \text{ M} \]. \

Step 3: 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. \

Step 4: 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 \(C\):\[ C = \frac{(5.50 \times 10^{-4})^2}{1.7 \times 10^{-4}} \approx 1.78 \times 10^{-3} \text{ M} \]. \

Key Concepts

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:

\[ K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]} \]This equation explains that \( K_a \) is the ratio of the concentration of dissociated ions to the concentration of the undissociated acid. A larger \( K_a \) indicates a stronger acid, meaning more ionization occurs. For formic acid, \( K_a = 1.7 \times 10^{-4} \), suggesting it is a weak acid since the value is small.

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:

• \( \text{pH} = -\log [\text{H}^+] \)

For example, given a \( \text{pH} \) of 3.26, you can reverse-calculate the \( [\text{H}^+] \) concentration as follows:

• Solve \( 3.26 = -\log [\text{H}^+] \).
• This leads to \( [\text{H}^+] = 10^{-3.26} \).
• Which simplifies to approximately \( 5.50 \times 10^{-4} \text{ M} \).

This calculation allows us to understand how many hydrogen ions are in the solution, and it's crucial for determining the acidity of solutions in chemistry.

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.

To analyze equilibrium, we often set up an expression with \( K_a \), considering both the concentrations of ions and undissociated acid:\[ K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]} \]Assuming that \( x \) is the equilibrium concentration of \( \text{H}^+ \) and \( \text{HCOO}^- \), and knowing \( x \) is quite small compared to the initial acid concentration \( C \), allows us to simplify and approximate solutions. This is vital for calculating unknown concentrations using equilibrium concepts.