Question

A buffered solution is made by adding 50.0g NH4Cl to 1.00 L of a 0.75M solution of NH3. Calculate the pH of the final solution. (assume no volume change).

Answer 1

The pH of the final solution after adding 50.0g NH4Cl to 1.00 L of a 0.75M solution of NH3 is calculated using the Henderson-Hasselbalch equation, yielding a pH of approximately 9.35.

Step-by-step explanation:

To calculate the pH of the final solution after adding 50.0g NH4Cl to 1.00 L of a 0.75M solution of NH3, we use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

First, we convert the mass of NH4Cl to moles:

Molar mass of NH4Cl = 53.49 g/mol

50.0 g NH4Cl * 1 mol/53.49 g = 0.935 moles NH4Cl

Since there is no volume change, the concentration of NH4+ from NH4Cl is 0.935 M. Now we have:

Concentration of NH4+ = 0.935 M (from NH4Cl)

Concentration of NH3 = 0.75 M (given)

The pKa of NH4+ is approximately 9.25. Applying the Henderson-Hasselbalch equation, we find the pH:

pH = 9.25 + log(0.935/0.75)

= 9.25 + log(1.247)

= 9.25 + 0.096

= 9.346

Therefore, the pH of the final solution is approximately 9.35.

Answer 2

9.15

Step-by-step explanation:

To solve this problem we'll use Henderson-Hasselbach's equation:

• pH = pKa + log
  `([A^-])/([HA])`

For this problem:

• pH = 9.25 + log
  `([NH_3])/([NH_4Cl])`

Now we calculate [NH₄Cl], first by converting 50.0 g of NH₄Cl into moles using its molar mass:

• 50.0 g NH₄Cl ÷ 53.491 g/mol = 0.935 mol

Meaning that [NH₄Cl] = 0.935 mol / 1.00 L = 0.935 M

Finally we calculate the pH:

• pH = 9.25 + log
  `(0.75M)/(0.935 M)` = 9.15