Question

Ammonia gas is bubbled into \(275 \mathrm{~mL}\) of water to make an aqueous solution of ammonia. To prepare a buffer with a \(\mathrm{pH}\) of \(9.56,15.0 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) are added. How many liters of \(\mathrm{NH}_{3}\) at \(25^{\circ} \mathrm{C}\) and 0.981 atm should be used to prepare the buffer? Assume no volume changes and ignore the vapor pressure of water.

Short answer

Answer: 14.45 L of ammonia gas is required to prepare the buffer solution.

Step-by-step solution

Calculate moles of NH4Cl

Given we have 15.0 g of NH₄Cl, we need to calculate the number of moles. We use the molecular weight of NH₄Cl which is approximately 53.49 g/mol. Moles of NH₄Cl = \(\frac{15.0 \mathrm{~g}}{53.49 \mathrm{~g/mol}} = 0.280\) mol

Use Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentrations of its acidic and basic components: pH = pKa + log(\(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\)) We're given pH = 9.56 and pKa of NH₄⁺ is 9.25, So, 9.56 = 9.25 + log(\(\frac{[\mathrm{NH₃}]}{[\mathrm{NH₄}⁺]}\)) Let \(r = \frac{[\mathrm{NH₃}]}{[\mathrm{NH₄}⁺]}\), then we can solve for r: log(r) = 9.56 - 9.25 r = 10^(0.31) = 2.045

Calculate moles of NH3

Using the ratio of NH₃ and NH₄⁺ concentrations we calculated (r = 2.045), we can find the moles of NH₃ required. 0.280 mol NH₄⁺ * r = 0.280 mol * 2.045 = 0.572 mol NH₃

Convert moles of NH3 to volume

Now we can use the ideal gas law to find the volume of NH₃ gas needed: PV = nRT Where P is the partial pressure (0.981 atm), n is the number of moles (0.572 mol), R is the gas constant (0.0821 L⋅atm/mol⋅K), and T is the temperature in Kelvin (25°C = 298 K). V = \(\frac{nRT}{P} = \frac{0.572 \mathrm{~mol} * 0.0821 \mathrm{~L\cdot atm/mol\cdot K} * 298 \mathrm{~K}}{0.981 \mathrm{~atm}}\) V = 14.45 L Thus, to prepare the buffer solution, 14.45 L of NH₃ gas is required at 0.981 atm and 25°C.