Question

A buffer is 0.10 M in NH3 and 0.10 M in NH4Cl. What is the pH of the solution after the addition of 10.0 mL of 0.20 M of HCl to 100.0 mL of the buffer?

Answer 1

Moles of NH₃:

0.1 M x 0.1 L = 0.01 moles

Moles of NH₄Cl:

0.1 M x 0.1 L = 0.01 moles

The addition of HCl changes the amounts of NH₃ and NH₄⁺

Moles of HCl is

0.2 M HCl x 0.010 L = 0.002 moles of HCl.

When H⁺ ions (from the HCl) react with NH₃, it will decrease NH₃ concentration, but will increase NH₄⁺ concentration.

So, moles of NH₃ = 0.010 – 0.002 = 0.008

So, moles of NH₄⁺ = 0.010 + 0.002 = 0.012

Hence, the concentrations

[NH₃] = 0.008 mole / (100 mL + 10 mL)

          = 0.008 mole / (0.1 L + 0.010 L)

          = 0.008/ (0.11 ) M

          = 0.073 M

[NH4+] = 0.012 mole / (0.11 L)

           = 0.11 M

Now, using the Henderson Hasselbalch equation

pOH = pKa + log([Salt]/[Base])

Ka of NH₃ = 1.81 x 10^-5

pKa = -log (1.81 x 10^-5) = 4.74

pOH = 4.74 + log (0.11 / 0.073)

        = 4.74 + log (1.51)

        =4.74 + 0.18 = 4.92

So, pH = 14 – 4.92 = 9.08