Finding pH of a buffer solution

Question

100 mL of buffer solution that is 0.15M HA (Ka= 6.8x10-5) and 0.20M NaA is mixed with 18.8mL of 0.25M HCl. What is the pH of the resulting solution?

Answer 1

Hello Linda

 

As Lauren mentioned, the answer is complicated, therefore, I will provide a very detailed answer in order for you to understand how to solve buffer problems like this.  My apologies for the lengthy answer!

 

A buffer is a mixture of a weak acid (HA) and its conjugate base (A^- or NaA expressed as a neutral salt).  The purpose of a buffer is to resist changes in the pH of the solution when either a small amount of a strong acid (H⁺) or base (OH^-) is added.  The components of the buffer solution, the weak acid and conjugate base, are written as an equilibrium reaction.

 

HA (aq) ⇔ H⁺ (aq) + A^- (aq)

 

When HCl, a strong acid, is added, will it react with HA or A^-?  Helpful hint:  Will an acid (HCl) react with another acid (HA)??  Of course not!  Therefore, HCl will react with A^- (a weak base) and cause the equilibrium reaction to shift towards the left → formation of HA.  NOTE!  Since HCl is a strong acid, it exists in solution as H⁺ and Cl^- ions.  Only H⁺ will participate in the reaction with A- as denoted below.

 

HA (aq) ← H⁺ (aq) + A^- (aq)

 

In other words, HCl (H⁺) will react with A^- to form more HA.  This will cause the concentration of A^- to decrease and the concentration of HA to increase.  Okay, let's do some math now.

 

Convert volume of the buffer and HCl from mL to Liters.

 

V_buffer = 100 mL x (1 L / 1000 mL) = 0.1 L

V_HCl = 18.8 mL x (1 L / 1000 mL) = 0.0188 L

 

Convert Molarity of HA, A^- and HCl to their respective # of moles.

 

Molarity = mol / L

 

mol = Molarity x L

 

mol_HA = 0.15 mol/L x 0.1 L = 0.015 mol HA

mol_A- = 0.20 mol/L x 0.1 L = 0.020 mol A-

mol_HCl = 0.25 mol/L x 0.0188 L = 0.0047 mol HCl

 

Notice that # of moles of HCl is significantly smaller than that of HA and A^-.  This means that 0.0047 mol of HCl will COMPLETELY react with A^- to form more HA.

 

mol_HA after HCl addition = 0.015 + 0.0047 = 0.0197 ≈ 0.020 mol HA

 

mol_A- after HCl addition = 0.020 - 0.0047 = 0.0153 ≈ 0.015 mol A^-

 

When HCl is added to the buffer solution, the final volume of the solution will increase.  V_buffer = 0.1 L + 0.0188 L = 0.1188 L

 

Next, calculate the Molarity of HA and A^- after the addition of HCl.

 

M = mol / L

 

M_HA = 0.020 mol / 0.1188 L = 0.17 M HA

 

M_A- = 0.015 mol / 0.1188 L = 0.13 M A^-

 

To calculate the pH of the buffer solution, we can use the Hendersen-Haaselbach equation.

 

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

 

K_a = 6.8 x 10^-5

pK_a = -log K_a = -log (6.8 x 10^-5) = 4.29

 

pH = 4.29 + log (0.13 / 0.17)

 

     = 4.29 + log (0.76)

 

     = 4.29 - 0.12

 

pH = 4.17

 

Hope the above was helpful!