Question

Calculate the \(\mathrm{pH}\) of a solution that is \(1.00 \mathrm{M} \mathrm{HNO}_{2}\) and \(1.00 \mathrm{M}\) \(\mathrm{NaNO}_{2}\)

Short answer

The pH of the solution of 1.00 M HNO₂ and 1.00 M NaNO₂ is approximately 3.35.

Step-by-step solution

Identify the initial concentrations

We are given a solution that is 1.00 M in both HNO₂ and NaNO₂. To calculate the pH, we need to consider the dissociation of these molecules in water. For the weak acid HNO₂, the dissociation reaction is: \[ HNO_{2}(aq) \rightleftharpoons H^{+}(aq) + NO_{2}^{-}(aq) \] For NaNO₂, the dissociation reaction is: \[ NaNO_{2}(aq) \rightarrow Na^{+}(aq) + NO_{2}^{-}(aq) \]

Determine the equilibrium concentrations

Since NaNO₂ is a strong electrolyte, it will dissociate completely in the solution, providing 1.00 M of NO₂⁻ ions initially. HNO₂, being a weak acid, will only partially dissociate. Let's call the extent of dissociation x. At equilibrium, the concentration of HNO₂ will be (1.00 - x) M, while the concentrations of H⁺ and NO₂⁻ will be x and (1+x) M, respectively.

Apply the Ka value of HNO₂

The dissociation constant Ka (the acid ionization constant) for HNO₂ can be found in a table of Ka values. The value for HNO₂ is: \[ K_{a} = 4.50 \times 10^{-4} \] Using the expression for the Ka, we can establish a relationship between the concentrations of the species at equilibrium: \[ K_{a} = \frac{[H^{+}][NO_{2}^{-}]}{[HNO_{2}]} \] Substituting the equilibrium concentrations, we get: \[ 4.50 \times 10^{-4} = \frac{(x)(1+x)}{(1-x)} \]

Solve for x

Since Ka is a small value, we can approximate by assuming x is very small compared to 1 and thus: \[ 4.50 \times 10^{-4} \approx \frac{x}{(1-x)} \approx \frac{x}{1} \] Solving for x, we find: \[ x \approx 4.50 \times 10^{-4} \]

Use the Henderson-Hasselbalch equation

Now that we have the concentrations of H⁺, HNO₂, and NO₂⁻ at equilibrium, we can use the Henderson-Hasselbalch equation to find the pH of the solution: \[ pH = pK_{a} + \log \frac{[A^{-}]}{[HA]} \] Where pKₐ = -log(Kₐ), and [A⁻] and [HA] are the equilibrium concentrations of the conjugate base and the weak acid, respectively. Plugging in the values: \[ pH = -\log(4.50 \times 10^{-4}) + \log \frac{(1+4.50 \times 10^{-4})}{(1-4.50 \times 10^{-4})} \]

Calculate the pH

Solve the equation to find the pH: \[ pH \approx 3.35 \] The pH of the solution of 1.00 M HNO₂ and 1.00 M NaNO₂ is approximately 3.35.

Key concepts

Weak Acid

In chemistry, acids are often categorized based on their ability to donate protons (H⁺ ions). A **weak acid** does not completely dissociate in a solution. Instead, it partially splits into its ions, establishing an equilibrium between the undissociated acid and the ions produced.
A classic example is **nitrous acid (HNO₂)**. When dissolved in water, not all HNO₂ molecules release H⁺ ions. This behavior contrasts with strong acids, like hydrochloric acid (HCl), which dissociate completely.
This partial dissociation makes pH calculations slightly more complex because you need to consider both the acid and its conjugate base at equilibrium. Understanding weak acids is crucial to mastering buffer solutions and equilibrium calculations.

• The degree of dissociation is represented by a dissociation constant, known as the acid ionization constant, Kₐ.
• Weak acids are significant in buffer solutions, where they help resist changes in pH.
• In calculations, we often solve for the concentration of dissociated H⁺ ions to find the solution's pH.

Henderson-Hasselbalch equation

The **Henderson-Hasselbalch equation** is a pivotal tool in chemistry for estimating the pH of buffer solutions. It links the pH of a solution to the concentration of a weak acid (HA) and its conjugate base (A⁻). The formula is:
\[ pH = pK_a + \log \frac{[A^-]}{[HA]} \]
This equation allows us to bypass more complex equilibrium calculations when the solution acts as a buffer, which contains both a weak acid and its conjugate base.
Using the equation requires knowing or calculating the **pKₐ**, which is the negative logarithm of the acid's dissociation constant (Kₐ). Understanding the relationship between the components in the buffer gives rise to practical applications, like adjusting the pH of pharmaceuticals or biological systems.

• The equation simplifies pH prediction in buffer solutions made from weak acids and their salts.
• It applies best when the concentration of the weak acid and its conjugate base are similar.
• Commonly used in biological and chemical applications, where maintaining a specific pH is crucial.

Dissociation Constant

A **dissociation constant (Kₐ)** is a measure of a compound's ability to dissociate into its ions and reflects the strength of an acid in a solution. For weak acids like HNO₂, Kₐ defines how readily the acid releases H⁺ ions.
The greater the Kₐ, the stronger the acid, implying a higher concentration of H⁺ ions at equilibrium. **Understanding Kₐ** is vital for calculating the pH of solutions containing weak acids, as it provides a direct link between the concentrations of the acid and its conjugate base:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
When using Kₐ in calculations, we often make approximations assuming that the change in the acid concentration due to dissociation is small compared to the initial concentration. This helps simplify the math in practical applications.

• Kₐ is specific to each acid and temperature-dependent.
• The pKₐ, or negative log of Kₐ, allows easy comparison of acid strengths.
• Knowing Kₐ is essential for determining equilibrium positions in acid-base reactions.