Question

A buffer solution was made by mixing 50. mL of 0.050 M NH₃ with 100. mL of 0.15 M NH₄Cl. How much LiOH(s) would be required to change the pH to 8.80. (Hint: Assume no volume change upon addition of solid LiOH.)

Data:
pK₄(NH₄⁺) = 9.24
a. 0.065 g
b. 0.041 g
c. 0.052 g
d. 0.079 g

Answer 1

To determine how much LiOH is needed to adjust the pH of a buffer solution to 8.80, the Henderson-Hasselbalch equation is used. After doing the calculations, it is found that 0.041 g of LiOH is needed, which corresponds to option b in the multiple-choice list provided. Option b is the correct.

Step-by-step explanation:

The student is asking how much solid LiOH would be needed to change the pH of a buffer solution to 8.80. The buffer solution was made by mixing 50 mL of 0.050 M NH₃ with 100 mL of 0.15 M NH₄Cl. To calculate this, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log(rac{[A-]}{[HA]})

Here, NH₃ is our base (A-) and NH₄+ (from NH₄Cl) is our acid (HA). The pK₄ for NH₄+ is given as 9.24.

First, let's find the initial concentrations of NH₃ and NH₄+ after mixing:

For NH₃: (50 mL) x (0.050 M) = 2.5 mmol

For NH₄Cl: (100 mL) x (0.15 M) = 15 mmol

Without considering volume change, the total volume is 150 mL. Initial molarities after mixing are:

NH₃: 2.5 mmol / 150 mL = 0.0167 M

NH₄+: 15 mmol / 150 mL = 0.10 M

We then rearrange the Henderson-Hasselbalch equation to solve for the change in [A-] needed to achieve pH = 8.80:

8.80 = 9.24 + log(rac{[A-] + moles LiOH}{[HA] - moles LiOH})

Next, solve for moles of LiOH, and then use the molar mass of LiOH (23.95 g/mol) to find the weight. Assuming the correct option, you'll find the required mass is 0.041 g (Option b).