Calculate the pH of the following buffer solution that resulted from 125ml of 0.1M NH3 with 250ml 0.1 M NH4Cl ( for NH3, Kb= 1.76x10^-5)

Use the Henderson-Hasselbalch equation. NH4Cl is the acid and NH3 is the base.

Does that mean that I do not need to do an ICE table but just use the HH equation?

To calculate the pH of a buffer solution, we need to use the Henderson-Hasselbalch equation:

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

In this case, NH3 acts as a base (A-) and NH4Cl acts as its conjugate acid (HA).

First, we need to calculate the concentrations of NH3 and NH4Cl in the final buffer solution.

For NH3:
Initial concentration = 0.1 M
Volume = 125 mL
Final concentration = (Initial concentration * Volume) / Total volume
= (0.1 M * 125 mL) / (125 mL + 250 mL)
= 0.0333 M

For NH4Cl:
Initial concentration = 0.1 M
Volume = 250 mL
Final concentration = (Initial concentration * Volume) / Total volume
= (0.1 M * 250 mL) / (125 mL + 250 mL)
= 0.0667 M

Next, we need to calculate the pKa of NH4Cl to determine the value of Ka.

pKa = -log(Ka)

Given that the Kb of NH3 is 1.76x10^-5, we can use the relationship:

Ka * Kb = Kw (the ion product constant for water)

Ka = Kw / Kb = (1.00x10^-14) / (1.76x10^-5)
= 5.68x10^-10

pKa = -log(5.68x10^-10) ≈ 9.244

Now we can substitute the values into the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])
= 9.244 + log(0.0333 M / 0.0667 M)

Calculating the ratio [A-]/[HA]:

[A-]/[HA] = 0.0333 M / 0.0667 M
= 0.5

Substituting the values into the equation:

pH = 9.244 + log(0.5)
≈ 9.244 + (-0.30)
≈ 8.94

Therefore, the pH of the buffer solution is approximately 8.94.