Question

A buffer is prepared by dissolving \(0.0250 \mathrm{~mol}\) of sodium nitrite, \(\mathrm{NaNO}_{2}\), in \(250.0 \mathrm{~mL}\) of \(0.0410 \mathrm{M}\) nitrous acid, \(\mathrm{HNO}_{2}\). Assume no volume change after \(\mathrm{HNO}_{2}\) is dissolved. Calculate the \(\mathrm{pH}\) of this buffer.

Short answer

Answer: The pH of the prepared buffer solution is approximately 3.737.

Step-by-step solution

Calculate the initial concentration of HNO2

First, we need to find the initial concentration of HNO2 in the solution. Given that its molarity is \(0.0410 \mathrm{M}\) and the volume of the solution is \(250.0 \mathrm{~mL}\), we can find the moles of HNO2. $$ [\mathrm{HNO}_{2}]_{initial} = 0.0410 \mathrm{M} $$ $$ moles\ of\ \mathrm{HNO}_{2} = [\mathrm{HNO}_{2}]_{initial}\times Volume\ of\ the\ solution $$ $$= 0.0410\ \mathrm{M} \times 0.250\ \mathrm{L} $$ $$= 0.01025\ \mathrm{mol} $$

Determine the moles of NaNO2 and NO2-

We know that the buffer solution contains \(0.0250 \mathrm{~mol}\) of sodium nitrite (NaNO2). Sodium nitrite will dissociate fully in the solution, producing an equal number of moles of NO2- ions: $$\mathrm{NaNO}_{2} \rightarrow \mathrm{Na}^{+} + \mathrm{NO}_{2}^-$$ Thus, moles of NO2- = \(0.0250\ \mathrm{mol}\)

Calculate the final concentrations of HNO2 and NO2-

Now, let's find the final concentrations of HNO2 and NO2- ions in the solution. We will assume that the volume of the solution does not change. Concentration of HNO2: $$[\mathrm{HNO}_{2}]_{final}=\frac{moles\ of\ \mathrm{HNO}_{2}}{Total\ volume\ of\ the\ solution}=\frac{0.01025\ \mathrm{mol}}{ 0.250\ \mathrm{L}}=0.0410\ \mathrm{M}$$ Concentration of NO2-: $$[\mathrm{NO}_{2}^-]_{final}=\frac{moles\ of\ \mathrm{NO}_{2}^-}{Total\ volume\ of\ the\ solution}=\frac{0.0250\ \mathrm{mol}}{ 0.250\ \mathrm{L}}=0.100\ \mathrm{M}$$

Use the Henderson-Hasselbalch equation

Now, we can apply the Henderson-Hasselbalch equation to determine the pH of the buffer: pH = pKa + log(\(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\)) We know that Ka for HNO2 is \(4.5 \times 10^{-4}\). To find the pKa, we use: pKa = -log(Ka) = -log(\(4.5 \times 10^{-4}\)) ≈ 3.35 Now we can plug the concentrations of the acid and base into the equation: pH = 3.35 + log(\(\frac{0.100\ \mathrm{M}}{0.0410\ \mathrm{M}}\)) ≈ 3.35 + log(2.439) ≈ 3.35 + 0.387 ≈ 3.737

Final answer

The final pH of the prepared buffer solution is approximately 3.737.