Question

A buffer solution is prepared by adding \(5.50 \mathrm{~g}\) of ammonium chloride and \(0.0188 \mathrm{~mol}\) of ammonia to enough water to make \(155 \mathrm{~mL}\) of solution. (a) What is the pH of the buffer? (b) If enough water is added to double the volume, what is the \(\mathrm{pH}\) of the solution?

Short answer

Answer: The pH of the initial buffer solution is 7.77, and after doubling the volume of the solution, the pH remains the same: 7.77.

Step-by-step solution

Find the concentration of ammonium chloride

To find the concentration of ammonium chloride, we first need to determine its molar mass, then we will use it to calculate the number of moles present in the solution and divide it by the volume of the solution. The molar mass of ammonium chloride is: NH4Cl = 14.01 (N) + 4 * 1.01 (H) + 35.45 (Cl) = 53.49 g/mol Now, we can find the number of moles in 5.50 g of ammonium chloride: moles(NH4Cl) = (5.50 g) / (53.49 g/mol) = 0.1027 mol The concentration of ammonium chloride in the buffer is: [NH4Cl] = (0.1027 mol) / (0.155 L) = 0.6626 mol/L

Find the concentration of ammonia

We were given the number of moles of ammonia added to the solution, which was 0.0188 mol. To find its concentration, we can simply divide the moles by the volume of the solution (0.155 L). The concentration of ammonia in the buffer is: [NH3] = (0.0188 mol) / (0.155 L) = 0.1213 mol/L

Calculate the pH of the buffer

Now that we have the concentrations of ammonium chloride and ammonia, we can calculate the pH of the buffer using the Henderson-Hasselbalch equation: \(\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\mathrm{NH3}]}{[\mathrm{NH4Cl}]}\) The \(K_\mathrm{a}\) of ammonia is \(5.56\times 10^{-10}\). To find \(\mathrm{p}K_\mathrm{a}\), we can take the negative logarithm of it: \(\mathrm{p}K_\mathrm{a}=-\log(5.56\times10^{-10})=9.25\) Now, using the equation, we can find the pH of the buffer solution: \(\mathrm{pH} =9.25 + \log\frac{0.1213}{0.6626} = 9.25 - 1.48 = 7.77\)

Calculate the concentrations after doubling the volume

When we double the volume of the solution, the concentrations of NH3 and NH4Cl will decrease, but their ratio will remain the same. We can find the new concentrations by dividing the initial concentrations by 2: New [NH3] = 0.1213 / 2 = 0.06065 mol/L New [NH4Cl] = 0.6626 / 2 = 0.3313 mol/L

Calculate the new pH after doubling the volume

We can use the Henderson-Hasselbalch equation again, with the new concentrations of NH3 and NH4Cl, to find the new pH: \(\mathrm{pH} = \mathrm{p}K_\mathrm{a}+\log\frac{[\mathrm{NH3}]}{[\mathrm{NH4Cl}]}\) \(\mathrm{pH} =9.25 + \log\frac{0.06065}{0.3313} = 9.25 - 1.48 = 7.77\) (a) The pH of the initial buffer solution is 7.77. (b) After doubling the volume of the solution, the pH remains the same: 7.77.