The weak base-conjugate acid buffer used in this laboratory consists of a weak base ammonia, NH3, and its conjugate acid ammonium chloride, NH4Cl. If the NH3 concentration is 0.05 M and the NH4Cl concentration is 0.05 M, what is the pH of the buffer? Kb for NH3 is 1.8*10^-5

Use the Henderson-Hasselbalch equation.

base = 0.05 M
acid = 0.05 M

To find the pH of the buffer solution, we need to consider the equilibrium between the weak base (NH3) and its conjugate acid (NH4Cl).

The balanced equation for the dissociation of NH4Cl is:
NH4Cl ⇌ NH4+ + Cl-

Since NH4Cl is a strong electrolyte, it will completely dissociate in water, and the concentration of NH4+ will be equal to the initial concentration of NH4Cl, which is 0.05 M.

Now, let's write the equation for the dissociation of NH3:
NH3 + H2O ⇌ NH4+ + OH-

The reaction can be considered to be in equilibrium because NH3 is a weak base. The concentration of OH- is equal to the concentration of NH4+ and can be calculated using the equilibrium constant expression:

Kb = [NH4+][OH-] / [NH3]

We can rearrange the equation to solve for [OH-]:

[OH-] = Kb * [NH3] / [NH4+]

Substituting the given values into the equation:
[OH-] = (1.8*10^-5) * (0.05) / (0.05)

[OH-] = 1.8*10^-5 M

Since the solution is basic (the concentration of OH- ions is higher than the concentration of H3O+ ions), we can calculate the pOH:

pOH = -log10 [OH-]
pOH = -log10 (1.8*10^-5)

pOH ≈ 4.74

Finally, we can calculate the pH using the equation:

pH = 14 - pOH
pH ≈ 14 - 4.74
pH ≈ 9.26

Therefore, the pH of the buffer is approximately 9.26.

To find the pH of the buffer, we need to determine how the weak base and its conjugate acid affect the solution's acidity or basicity.

The reaction that occurs in this buffer solution can be represented as follows:

NH3 + H2O ⇌ NH4+ + OH-

First, let's calculate the concentration of OH- by using the Kb (base dissociation constant) for NH3. The Kb expression for this reaction is written as:

Kb = [NH4+][OH-] / [NH3]

Given that the NH3 concentration is 0.05 M, and Kb for NH3 is 1.8*10^-5, we can calculate the OH- concentration:

1.8*10^-5 = (0.05-x)(x) / (0.05)

Since x represents the concentration of OH-, the change in NH4+ concentration can be neglected compared to the initial concentration of NH3 because the reaction lies to the left. Thus, we can assume that the (0.05 - x) term is approximately equal to 0.05.

1.8*10^-5 = (0.05)(x) / (0.05)

Rearranging the equation:
x = (1.8*10^-5)(0.05) / (0.05)

Simplifying:
x = 1.8*10^-5

Now we have the concentration of OH-, which means we can calculate the pOH:

pOH = -log10(OH-) = -log10(1.8*10^-5)

pOH ≈ 4.74

Since pH + pOH = 14, we can find the pH of the buffer solution:

pH ≈ 14 - 4.74

pH ≈ 9.26

Therefore, the pH of the buffer solution is approximately 9.26.