Question

To prepare a buffer properly, it is essential to understand how a buffer is described. Here are four buffers used:

1. A solution that is 0.1 M HC2H3O2 and 0.10 M NaC2H3O2

2. One part 0.10 M NH3 mixed with one part 0.10 M NH4Cl

3. A solution that is 0.10 M NH3 and 0.20 M NH4Cl

4. Three parts 0.10 M NH3 mixed with one part 0.10 M HCl

For the first and third buffers, the concentrations of each component are explicit. What are the identities of and concentrations of the buffer components in the second and third buffers? (Keep in mind when mixed, the original solutions dilute each other)

Buffer 2:

Buffer 4:

Answer 1

2. One part 0.10 M NH3 mixed with one part 0.10 M NH4Cl

Say each part is about 100.0 mL = 100mL*1L/1000mL = 0.100 L

When 1:1 parts of NH₃ and NH₄Cl are mixed, total solution volume = 200.0 mL*1L/1000mL = 0.200 L

Moles of NH₃ =  0.10 mol/L * 0.100 L = 0.010 mol

Molarity of NH₃ after mixing = 0.010 mol /0.200 L = 0.05 M

Moles of NH₄Cl =  0.10 mol/L * 0.100 L = 0.010 mol

Molarity of NH₄Cl after mixing = 0.010 mol /0.200 L = 0.05 M

pK_a of NH₄⁺ = 14.00 - pK_b(NH3)

   = 14.00 - 4.75

= 9.25

= 9.25

4. Three parts 0.10 M NH₃ mixed with one part 0.10 M HCl

If one part is 100.0 mL

Three parts will be 300.0 mL

When 3:1 parts of NH₃ and HCl are mixed, total solution volume = 400.0 mL*1L/1000mL = 0.400 L

Moles of NH₃ =  0.10 mol/L * 0.300 L = 0.030 mol

Moles of HCl =  0.10 mol/L * 0.100 L = 0.010 mol

NH₃   + HCl -----> NH₄Cl

0.030 mol 0.010 mol 0

-0.010 mol     -0.010 mol +0.010 mol

Moles of NH₃ = 0.030 mol - 0.010 mol = 0.020 mol

Molarity of NH₃ =0.020 mol/0.400 L = 0.05 M

Moles of NH₄⁺ = 0.010 mol

Molarity of NH₄⁺ = 0.010 mol/0.400 L = 0.025 M

= 9.55