Question

What volumes of 0.50 \(\mathrm{M} \mathrm{HNO}_{2}\) and 0.50 \(\mathrm{M}\) NaNO, must be mixed to prepare 1.00 \(\mathrm{L}\) of a solution buffered at \(\mathrm{pH}=3.55 ?\)

Short answer

To prepare 1.00 L of a solution buffered at pH 3.55, we need to mix approximately \(0.723\mathrm{L}\) of 0.50 M HNO₂ and \(0.277\mathrm{L}\) of 0.50 M NaNO₂.

Step-by-step solution

Calculate the ionization constant of HNO₂

First, let's find the ionization constant (Ka) for HNO₂ (a weak acid). We can use the given pH value to determine the hydrogen ion concentration (\([H^+]\)): \[ \mathrm{pH} = -\log([H^+]), \] Rearranging this equation for \([H^+]\): \[ [H^+] = 10^{-\mathrm{pH}}. \] Plugging in the given pH value of 3.55: \[ [H^+] = 10^{-3.55} \approx 2.818 \times 10^{-4} \mathrm{M}. \] Next, we know that HNO₂ ionizes according to the following equation: \[ \mathrm{HNO_2} \rightleftharpoons \mathrm{H^+} + \mathrm{NO_2^-}. \] The Ka expression for this reaction is: \[ K_\mathrm{a} = \frac{[\mathrm{H^+}][\mathrm{NO_2^-}]}{[\mathrm{HNO_2}]}. \] Given that the initial concentrations of HNO₂ and NaNO₂ are equal (0.50 M), we can set the dissociated concentration of HNO₂ to 0.5 M - x and the concentration of NO₂⁻ to x, so the equation becomes: \[ K_\mathrm{a} = \frac{([H^+] + x)x}{(0.5 - x)}. \]

Use the Henderson-Hasselbalch equation

Now, we will use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa and concentrations of acid and base: \[ \mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\mathrm{base}]}{[\mathrm{acid}]}, \] where base is the concentration of NO₂⁻ and acid is the concentration of HNO₂. We can rearrange this equation to find the ratio of base concentration to acid concentration: \[ \frac{[\mathrm{base}]}{[\mathrm{acid}]} = 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}. \]

Calculate the volume ratio

Let V₁ be the volume of HNO₂ and V₂ be the volume of NaNO₂. We can relate the volumes to the concentrations: \[ \frac{[\mathrm{base}]}{[\mathrm{acid}]} = \frac{0.50 \mathrm{M} \times V₂}{0.50 \mathrm{M} \times V₁} = \frac{V₂}{V₁}. \] Now, we can substitute the value derived in Step 2 into this equation: \[ \frac{V₂}{V₁} = 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}, \] where \(\mathrm{pH} = 3.55\) and \(K_\mathrm{a} = 4.5 \times 10^{-4}\). The total volume is 1.00 L, so we have the following equation for V₁ and V₂: \[ V₁ + V₂ = 1.00\mathrm{L}. \]

Solve for the volumes

Solve the system of equations established in Steps 3: \[ 1. \; V₁ + V₂ = 1.00\mathrm{L}, \] \[ 2. \; \frac{V₂}{V₁} = 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}. \] From equation 2: \[ V₂ = 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}V₁, \] Substitute this expression for V₂ into equation 1: \[ V₁ + 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}V₁ = 1.00\mathrm{L}, \] \[ V₁(1 + 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}) = 1.00\mathrm{L}. \] Then, isloate V₁: \[ V₁ = \frac{1.00\mathrm{L}}{1 + 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}}. \] Now we can plug in the given pH value and calculated pKa: \[ V₁ \approx 0.723\mathrm{L}. \] Then, we can find V₂ by subtracting V₁ from the total volume (1.00 L): \[ V₂ = 1.00\mathrm{L} - V₁ \approx 0.277\mathrm{L}. \] Thus, to prepare 1.00 L of a solution buffered at pH 3.55, we need to mix approximately 0.723 L of 0.50 M HNO₂ and 0.277 L of 0.50 M NaNO₂.

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a vital formula in chemistry used for calculating the pH of a buffer solution. A buffer is a mixture that minimizes pH changes, even with the addition of acids or bases.
The equation is derived from the equilibrium expression for a weak acid:\[pH = pK_a + \log\left(\frac{[\text{base}]}{[\text{acid}]\right)\]Where:

• \(pH\) is the measure of acidity or basicity of the solution.
• \(pK_a\) is the negative base-10 logarithm of the acid dissociation constant \(K_a\).
• \([\text{base}]\) and \([\text{acid}]\) are the molar concentrations of the base and acid components of the buffer.

This formula helps compute how the concentrations of the components (acid and conjugate base) affect the pH.
By rearranging the equation, we can find the necessary concentrations of each component to achieve a desired pH.

Acid-Base Equilibrium

Acid-base equilibrium refers to the balance that occurs in a solution when an acid and its conjugate base are both present. This equilibrium is critical in the formation of a buffer solution.
For weak acids like HNO₂, they only partially ionize in water:\[\text{HNO}_2 \rightleftharpoons \text{H}^+ + \text{NO}_2^-\]The equilibrium helps buffers maintain a steady pH, as they resist significant changes even when small amounts of acid or base are added.
In this equilibrium:

• Weak acids don't completely dissociate, leaving a mix of ions and un-ionized molecules.
• The conjugate base, in this case, \(\text{NO}_2^-\), pairs with the weak acid to define the buffer capacity.

Understanding this equilibrium helps in designing effective buffers, crucial in many chemical and biological processes.

Ionization Constant (Ka)

The ionization constant \(K_a\) is a measure of the strength of an acid in solution. It indicates how well an acid breaks apart into its ions. For weak acids such as HNO₂, the \(K_a\) value is pivotal in predicting how much it ionizes in water.
The ionization of HNO₂ is represented by the equation:\[K_a = \frac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]}\]Where:

• \([\text{H}^+]\) and \([\text{NO}_2^-]\) are the concentrations of the ions formed.
• \([\text{HNO}_2]\) is the concentration of the original acid.

A larger \(K_a\) value indicates a stronger acid, which dissociates more completely.
For the problem at hand, knowing \(K_a\) allows us to calculate the necessary concentrations to achieve a desired pH using the Henderson-Hasselbalch equation.