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

The Henderson-Hasselbalch equation is given by: pH = pKa + log(\(\frac{[A^-]}{[HA]}\)), where pH is the desired pH of the buffer solution, pKa is the acid dissociation constant of the weak acid, [A⁻] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

The pKa of a weak acid can be calculated using the equation: pKa = -log(Ka), where Ka is the acid dissociation constant for HNO₂. For HNO₂, the Ka is 4.5 × 10⁻⁴. Using this value, we can calculate the pKa: pKa = -log(4.5 × 10⁻⁴) ≈ 3.35.

Now we can use the Henderson-Hasselbalch equation to find the ratio of the concentrations of the acid and its conjugate base: 3.55 = 3.35 + log(\(\frac{[A^-]}{[HA]}\)). Solving for the ratio of the concentrations, we get: \(\frac{[A^-]}{[HA]}\) = 10^(3.55 - 3.35) ≈ 1.585.

Let x be the volume (in L) of 0.50 M HNO₂ and y be the volume (in L) of 0.50 M NaNO₂ to be mixed. We need to satisfy two conditions: 1. The total volume of the solution must be 1.00 L, so x + y = 1.00. 2. The ratio of the concentrations of NaNO₂ to HNO₂ in the final solution must be 1.585. Since the volume of each solution contributes to the concentration of NaNO₂, we can rewrite the second condition with the concentrations and volumes of the two solutions: \(\frac{0.50y}{0.50x}\) = 1.585. Solving this equation, we find that y/x = 1.585. Using the first condition (x + y = 1.00), we can now solve for x and y: x = 1.00 – y, y = 1.585x. Now substitution x as 1.00 - y in the second equation: y = 1.585(1.00 - y). Solving for y, we get: y ≈ 0.614 L. Now substituting this value back into x: x = 1.00 - 0.614 ≈ 0.386 L.

The volumes of 0.50 M HNO₂ and 0.50 M NaNO₂ that must be mixed to prepare 1.00 L of the solution buffered at pH = 3.55 are approximately 0.386 L (or approx. 386 mL) of HNO₂ and 0.614 L (or approx. 614 mL) of NaNO₂.