Chapter 16: Problem 55
Calculate the \(\mathrm{pH}\) of an aqueous solution at \(25^{\circ} \mathrm{C}\) that is \(0.095 M\) in hydrocyanic acid \((\mathrm{HCN}) .\left(K_{\mathrm{a}}\right.\) for hydrocyanic acid \(\left.=4.9 \times 10^{-10} .\right)\)
Short Answer
Expert verified
The pH of the solution is approximately 5.17.
Step by step solution
01
Write the Dissociation Equation
Hydrocyanic acid (HCN) dissociates in water as follows: \[ \text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^- \]
02
Define Initial Concentrations
Initially, the concentration of HCN is given as 0.095 M. The concentrations of \(\text{H}^+\) and \(\text{CN}^-\) are both 0 M since the dissociation has not yet occurred.
03
Set Up the ICE Table
Using an ICE (Initial, Change, Equilibrium) table helps track the changes in concentrations. For the dissociation: \[ \begin{array}{c|c|c|c} & \text{HCN} & \text{H}^+ & \text{CN}^- \ \hline \text{Initial (M)} & 0.095 & 0 & 0 \ \text{Change (M)} & -x & +x & +x \ \text{Equilibrium (M)} & 0.095-x & x & x \ \end{array} \]
04
Write the Ka Expression
The expression for the acid dissociation constant \(K_a\) for HCN is: \[ K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} = \frac{x \cdot x}{0.095 - x} \] Given \(K_a = 4.9 \times 10^{-10}\), we plug in the values.
05
Make an Approximation
Since \(K_a\) is very small, we assume \(x \ll 0.095\), so \(0.095 - x \approx 0.095\). This simplifies the expression to: \[ 4.9 \times 10^{-10} = \frac{x^2}{0.095} \]
06
Solve for x
Multiply both sides by 0.095 to get: \[ x^2 = 4.9 \times 10^{-10} \times 0.095 \] Then calculate: \[ x^2 = 4.655 \times 10^{-11} \] Take the square root: \[ x = \sqrt{4.655 \times 10^{-11}} \approx 6.82 \times 10^{-6} \] This x is \([\text{H}^+]\).
07
Calculate pH
The \(\text{pH}\) is calculated using the formula \(\text{pH} = -\log[\text{H}^+]\). Therefore: \[ \text{pH} = -\log(6.82 \times 10^{-6}) \approx 5.17 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Acid Dissociation Constant
The acid dissociation constant, often represented as \( K_a \), is a crucial concept in chemistry. It reflects the strength of a weak acid in its ability to dissociate into ions in a solution. The higher the \( K_a \), the stronger the acid, as it indicates a greater degree of ionization.
For hydrocyanic acid (HCN), the given \( K_a \) is \( 4.9 \times 10^{-10} \). This small \( K_a \) value signifies that HCN is a weak acid. In other words, it does not dissociate completely in water. Understanding the \( K_a \) value helps predict how much of the acid will release hydrogen ions \( (H^+) \) when dissolved. This is fundamental when calculating the pH of weak acids.
For hydrocyanic acid (HCN), the given \( K_a \) is \( 4.9 \times 10^{-10} \). This small \( K_a \) value signifies that HCN is a weak acid. In other words, it does not dissociate completely in water. Understanding the \( K_a \) value helps predict how much of the acid will release hydrogen ions \( (H^+) \) when dissolved. This is fundamental when calculating the pH of weak acids.
ICE Table
An ICE Table (Initial, Change, Equilibrium Table) is a strategic tool in chemistry for solving equilibrium problems. It tracks the changes in concentration of each species in a chemical reaction.
In the case of HCN, the dissociation reaction is set up in the table as follows:
In the case of HCN, the dissociation reaction is set up in the table as follows:
- Initial concentrations: \( [HCN] = 0.095 \, M \), \( [H^+] = 0 \, M \), \( [CN^-] = 0 \, M \).
- Change in concentration: As the reaction proceeds, HCN concentration decreases by \( x \), while \( H^+ \) and \( CN^- \) increase by \( x \).
- Equilibrium concentrations: \( [HCN] = 0.095 - x \), \( [H^+] = x \), \( [CN^-] = x \).
Ka Expression
The \( K_a \) expression is a formula used to relate the concentrations of the products and reactants involved in an acid's dissociation reaction at equilibrium. For hydrocyanic acid, this expression is:\[ K_a = \frac{[H^+][CN^-]}{[HCN]} \]This equation represents the ratio of the concentration of dissociated ions to the undissociated acid. Inserting the ICE table values results in:\[ K_a = \frac{x \cdot x}{0.095 - x} \]Because \( K_a \) for HCN is so small, we assume \( x \) (the amount of ionization) is very small compared to the initial concentration, allowing us to approximate \( 0.095 - x \approx 0.095 \). This simplification makes it easier to solve for \( x \), which represents the hydrogen ion concentration \([H^+]\).
Hydrocyanic Acid
Hydrocyanic acid (HCN) is a weak acid, meaning it does not completely dissociate in water. This is important when calculating pH because only a small fraction releases hydrogen ions into the solution.
Its weak nature is evident from the low \( K_a \) value \(4.9 \times 10^{-10}\), which implies a limited formation of \( H^+ \) and \( CN^- \) ions. When solving pH problems involving HCN, it's essential to use assumptions that simplify the math, such as neglecting \( x \) in the denominator of the expression \( 0.095 - x \approx 0.095 \).
This simplification leads us to find the concentration of \( H^+ \), which then allows for easy calculation of the pH using the formula: \( \text{pH} = -\log[H^+] \). With these calculations, students can better understand the behavior of hydrocyanic acid in aqueous solutions.
Its weak nature is evident from the low \( K_a \) value \(4.9 \times 10^{-10}\), which implies a limited formation of \( H^+ \) and \( CN^- \) ions. When solving pH problems involving HCN, it's essential to use assumptions that simplify the math, such as neglecting \( x \) in the denominator of the expression \( 0.095 - x \approx 0.095 \).
This simplification leads us to find the concentration of \( H^+ \), which then allows for easy calculation of the pH using the formula: \( \text{pH} = -\log[H^+] \). With these calculations, students can better understand the behavior of hydrocyanic acid in aqueous solutions.