Calculating Changes in a Buffer Solution



 

Learning Objective

  • Calculate the final pH of a solution generated by the addition of a strong acid or base to a buffer.

Key Points

    • If the concentrations of the weak acid and its conjugate base in a buffer solution are reasonably high, then the solution is resistant to changes in hydrogen ion concentration, or pH.
    • The change in pH of a buffer solution with an added acid or base can be calculated by combining the balanced equation for the reaction and the equilibrium acid dissociation constant (Ka).
    • Comparing the final pH of a solution with and without the buffer components shows the effectiveness of the buffer in resisting a change in pH.

Terms

  • acid dissociation constantQuantitative measure of the strength of an acid in solution; typically written as a ratio of the equilibrium concentrations of products to reactants.
  • pHThe negative of the logarithm to base 10 of the concentration of hydrogen ions, measured in moles per liter; a measure of acidity or alkalinity of a substance, which takes numerical values from 0 (maximum acidity) through 7 (neutral) to 14 (maximum alkalinity).

If the concentrations of a solution of a weak acid and its conjugate base are reasonably high, then the solution is resistant to changes in hydrogen ion concentration. These solutions are known as buffers. It is possible to calculate how the pH of the solution will change in response to the addition of an acid or a base to a buffer solution.

Calculating Changes in a Buffer Solution, Example 1:

A solution is 0.050 M in acetic acid (HC2H3O2) and 0.050 M NaC2H3O2. Calculate the change in pH when 0.001 mole of hydrochloric acid (HCl) is added to a liter of solution, assuming that the volume increase upon adding the HCl is negligible. Compare this to the pH if the same amount of HCl is added to a liter of pure water.

Step 1:

[latex]{HC_2H_3O_2}(aq)\leftrightharpoons {H^+}(aq)+{C_2H_3O_2^-}(aq) [/latex]

Recall that sodium acetate,
NaC2H3O2, dissociates into its component ions, Na+ and C2H3O2 (the acetate ion) upon dissolution in water. Therefore, the solution will contain both acetic acid and acetate ions.

Before adding HCl, the acetic acid equilibrium constant is:

[latex]{ K }_{ a }=\frac { { [H }^{ + }]{ [C_2{H}_3{O}_2 }^{ - }] }{ [H{C}_2{H}_3{O}_2] } =\frac { x(0.050) }{ (0.050) }[/latex]

(assuming that x is small compared to 0.050 M in the equilibrium concentrations)

Therefore:

[latex]x=[H^+]={ K }_{ a }=1.76\times 1{ 0 }^{ -5 }M[/latex]

[latex]pH={ pK }_{ a }=4.75[/latex]

In this example, ignoring the x in the [C2H3O2] and [HC2H3O2] terms was justified because the value is small compared to 0.050.

Step 2:

The added protons from HCl combine with the acetate ions to form more acetic acid:

[latex]C_2H_3O_2^{ - }+{ H }^{ + }(from HCl)\rightarrow HC_2H_3O_2[/latex]

Since all of the H+ will be consumed, the new concentrations will be [latex][HC_2H_3O_2]=0.051 M[/latex] and [latex][C_2H_3O_2^-]=0.049 M[/latex] before the new equilibrium is to be established. Then, we consider the equilibrium conentrations for the dissociation of acetic acid, as in Step 1:

[latex]{HC_2H_3O_2}(aq)\leftrightharpoons {H^+}(aq)+{C_2H_3O_2^-}(aq) [/latex]

we have,

[latex]{ K }_{ a }=\frac { x(0.049) }{ (0.051) }[/latex]

[latex]x=[H^+]=(1.76\times 1{ 0 }^{ -5 })\frac { 0.051 }{ 0.049 } =1.83\times 1{ 0 }^{ -5 }M[/latex]

[latex]pH=-log([{ H }^{ + }])=4.74[/latex]

In the presence of the acetic acid-acetate buffer system, the pH only drops from 4.75 to 4.74 upon addition of 0.001 mol of strong acid HCl, a difference of only 0.01 pH unit.

Step 3:

Adding 0.001 M HCl to pure water, the pH is:

[latex]pH=-log([{ H }^{ + }])=3.00[/latex]

In the absence of HC2H3O2 and C2H3O2, the same concentration of HCl would produce a pH of 3.00.

Calculating Changes in a Buffer Solution, Example 2:

A formic acid buffer is prepared with 0.010 M each of formic acid (HCOOH) and sodium formate (NaCOOH). The Ka for formic acid is 1.8 x 10-4. What is the pH of the solution? What is the pH if 0.0020 M of solid sodium hydroxide (NaOH) is added to a liter of buffer? What would be the pH of the sodium hydroxide solution without the buffer? What would the pH have been after adding sodium hydroxide if the buffer concentrations had been 0.10 M instead of 0.010 M?

Step 1:

Solving for the buffer pH:

[latex]HCOOH \leftrightharpoons {H^+} + {HCOO^-}[/latex]

Assuming x is negligible, the Ka expression looks like:

[latex]{ K }_{ a }=\frac { x(0.010) }{ (0.010) }[/latex]

1.8 x 10-4 = x = [H+]

pH = -log [H+] = 3.74

Buffer: pH = 3.74

Step 2:

Solving for the buffer pH after 0.0020 M NaOH has been added:

[latex]OH^- + HCOOH \rightarrow {H_2O} + {HCOO^-}[/latex]

The concentration of HCOOH would change from 0.010 M to 0.0080 M and the concentration of HCOO would change from 0.010 M to 0.0120 M.

[latex]{ K }_{ a }=\frac { x(0.0120) }{ (0.0080) }[/latex]

After adding NaOH, solving for [latex]x=[H^+][/latex] and then calculating the pH = 3.92. The pH went up from 3.74 to 3.92 upon addition of 0.002 M of NaOH.

Step 3:

Solving for the pH of a 0.0020 M solution of NaOH:

pOH = -log (0.0020)

pOH = 2.70

pH = 14 – pOH

pH = 11.30

Without buffer: pH = 11.30

Step 4:

Solving for the pH of the buffer solution if 0.1000 M solutions of the weak acid and its conjugate base had been used and the same amount of NaOH had been added:

The concentration of HCOOH would change from 0.1000 M to 0.0980 M and the concentration of HCOO would change from 0.1000 M to 0.1020 M.

[latex]{ K }_{ a }=\frac { x(0.1020) }{ (0.0980) }[/latex]

pH if 0.1000 M concentrations had been used = 3.77

This shows the dramatic effect of the formic acid-formate buffer in keeping the solution acidic in spite of the added base. It also shows the importance of using high buffer component concentrations so that the buffering capacity of the solution is not exceeded.