Question

Calculate the $\mathrm{pH}$ of an aqueous solution at ${25}^{\circ }\mathrm{C}$ that is $0.095M$ in hydrocyanic acid $(\mathrm{HCN}).(K_{\mathrm{a}}$ for hydrocyanic acid $=4.9\times {10}^{-10}.)$

Short answer

The pH of the solution is approximately 5.17.

Step-by-step solution

Write the Dissociation Equation

Hydrocyanic acid (HCN) dissociates in water as follows: $$\text{HCN}\rightleftharpoons {\text{H}}^{+}+{\text{CN}}^{-}$$

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.

Set Up the ICE Table

Using an ICE (Initial, Change, Equilibrium) table helps track the changes in concentrations. For the dissociation: $$\text{Misplaced hline}$$

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.

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: $$4.9\times {10}^{-10}=\frac{x^2}{0.095}$$

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 $[{\text{H}}^{+}]$.

Calculate pH

The $\text{pH}$ is calculated using the formula $\text{pH}=-\log [{\text{H}}^{+}]$. Therefore: $$\text{pH}=-\log (6.82\times {10}^{-6})\approx 5.17$$

Key concepts

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.

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:

• Initial concentrations: $[HCN]=0.095M$, $[H^{+}]=0M$, $[C{N}^{-}]=0M$.
• Change in concentration: As the reaction proceeds, HCN concentration decreases by $x$, while $H^{+}$ and $C{N}^{-}$ increase by $x$.
• Equilibrium concentrations: $[HCN]=0.095-x$, $[H^{+}]=x$, $[C{N}^{-}]=x$.

This setup allows us to easily plug into the $K_a$ expression and solve for the unknowns.

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:$$K_a=\frac{[H^{+}][C{N}^{-}]}{[HCN]}$$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 $C{N}^{-}$ 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.