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 \(\mathrm{pH}=3.55 ?\)

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: \[ K_\mathrm{a} = \frac{([H^+] + x)x}{(0.5 - x)}. \]

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: \[ \frac{[\mathrm{base}]}{[\mathrm{acid}]} = 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}. \]

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 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.