Question

a. Calculate the pH of a buffer that is prepared by completely dissolving 0.0401 g of NH4Cl in exactly 25.0 ml of a 0.020 Molar aqueous NH3 solution.

b. What is the pH after 5.0 mL of 0.020 M NaOH solution is added to 0.020 L of this buffer?

Answer 1

A) The moles of the buffer components are calculated:

n NH4 + = g / MM = 0.0401 g / 18 g / mol = 0.0022 mol

[NH4 +] = 0.0022 / 0.025 = 0.088 M

n NH3 = M * V = 0.02 M * 0.025 L = 0.0005 mol

The pH is calculated:

pH = pKa + log n NH3 / n NH4 + = 9.25 + log (0.0005 / 0.0022) = 8.61

B) The moles of NaOH and the components of the buffer are calculated:

n NaOH = 0.02 M * 0.005 L = 0.0001 mol

n NH3 = 0.02 * 0.02 = 0.0004 mol

n NH4 + = 0.088 M * 0.02 L = 0.0018 mol

NaOH reacts with NH4 + and forms NH3, the new pH is calculated:

pH = 9.25 + log (0.0004 + 0.0001 / 0.0018 - 0.0001) = 8.72