What is the pH of buffer solution prepared with 0.05 M ammonia & 0.05 M ammonium chloride? The Kb value of ammonia is 1.80 ✖️ 10^-5 at 25C.

4.74

To find the pH of the buffer solution prepared with 0.05 M ammonia (NH3) and 0.05 M ammonium chloride (NH4Cl), we can use the Henderson-Hasselbalch equation:

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

In this case, ammonia (NH3) acts as the base (A-) and ammonium chloride (NH4Cl) acts as the acid (HA). The pKa can be calculated using the pKa = pKw - pKb equation, where pKw is the negative logarithm of the autoionization constant for water (1 x 10^-14) at 25°C.

First, let's find the pKb:

pKb = -log(Kb)

pKb = -log(1.80 x 10^-5)
pKb = -(-4.75)
pKb = 4.75

Now, we can calculate the pKa:

pKa = 14 - pKb
pKa = 14 - 4.75
pKa = 9.25

Next, we can substitute the values into the Henderson-Hasselbalch equation:

pH = 9.25 + log([A-]/[HA])

Since the concentrations of ammonia and ammonium chloride are both 0.05 M, [A-]/[HA] = 1.

pH = 9.25 + log(1)
pH = 9.25

Therefore, the pH of the buffer solution is 9.25.

To find the pH of a buffer solution prepared with ammonia (NH3) and ammonium chloride (NH4Cl), we need to consider the equilibrium between ammonia and ammonium ion (NH4+). The reaction is as follows:

NH3 + H2O ⇌ NH4+ + OH-

Since we are given the concentration of ammonia and ammonium chloride, we can assume that they are fully ionized, resulting in an equal concentration of NH4+ and OH- ions. Therefore, in the buffer solution, the concentration of hydroxide ions (OH-) will be equal to the concentration of ammonium ions (NH4+), which can be calculated using the Kb value of ammonia.

The Kb expression for ammonia is as follows:
Kb = [NH4+][OH-] / [NH3]

We can rearrange the equation to solve for [OH-]:
[OH-] = Kb * [NH3] / [NH4+]

Let's substitute the given values into the equation:
Kb = 1.80 * 10^-5
[NH3] = 0.05 M
[NH4+] = 0.05 M

Now, we can calculate the concentration of hydroxide ions ([OH-]):
[OH-] = (1.80 * 10^-5) * (0.05) / (0.05) = 1.80 * 10^-5

Since the concentration of hydroxide ions is equal to the concentration of ammonium ions in the buffer solution, we can calculate the pOH:

pOH = -log[OH-] = -log(1.80 * 10^-5) ≈ 4.74

Finally, we can find the pH of the buffer solution by subtracting the pOH from 14:

pH = 14 - pOH ≈ 14 - 4.74 ≈ 9.26

Therefore, the pH of the buffer solution prepared with 0.05 M ammonia and 0.05 M ammonium chloride is approximately 9.26.