Buffer Effectiveness, Titrations & pH curves

Section 16.3-16.4 Buffer effectiveness

Buffer effectiveness refers to the ability of a buffer to resist pH change

Effective buffers only neutralize small to moderate amounts of acid or base

Factors influencing buffer effectiveness:

● Ratio of the buffers acid-conjugate base concentrations ○ Composition of buffer solution ● Overall absolute concentration ○ Concentration of buffer solution

Optimal conditions for buffer effectiveness can be derived from the Henderson-Hasselbalch equation Reference Table Relative amounts of Acid and Base

Buffers are most effective when their acid and conjugate base concentrations are equal

Buffers become less effective as the difference between the concentrations of acid and conjugate base increase

In order for a buffer to be effective the difference in concentration should not differ by more than a factor of 10 The math

This can be illustrated by examining two solutions of a generic buffer, pKa=5.00, neutralizing 0.01 mol NaOH.

Both solutions have a volume of 1.0 liter and 0.20 mol total acid and conjugate base

Solution I has equal concentrations of acid and conjugate base while Solution II has 0.18 mol acid (HA) and 0.02 mol base (A-). Solution I Solution II Initial pH =5.00 + log (1.00) = 5.00 Initial pH =5.00 + log (0.02/0.18) = 4.05 - ⇒ - - ⇒ - OH + HA H2O + A OH + HA H2O + A

Before 0.00 0.10 mol 0.10mol Before 0.00 0.18 mol 0.02 mol

Addition 0.01 mol — — Addition 0.01 mol — —

After 0.00 mol 0.09 mol .110 mol After 0.00 mol 0.17 mol .03 mol

pH =pK + log ( [base] / [acid] ) pH =pK + log ( [base] / [acid] ) a a Reference Table =5.00 + log ( 0.1100 / .090 ) =5.00 + log ( 0.03 / .17 ) =5.09 =4.25 % change=(5.09 - 5.00) / 5.00 X100% % change=(4.25 - 4.05) / 4.05 X100% =1.8% =5.0% In the context of the H.-H. equation

pH=pKa + log ( [base] / [acid] )

[base] = [acid] ⇒ [base] / [acid] = 1 log(1)=0

pH=pKa + 0 pH=pKa Concentrations of Acid and Base

Buffers are most effective at high concentrations of acid and conjugate base

The more dilute the buffer components the less effective Solution I Solution II Initial pH =5.00 + log (0.50/0.50) = 5.00 Initial pH =5.00 + log (0.050/0.050) = 5.00 - ⇒ - - ⇒ - OH + HA H2O + A OH + HA H2O + A

Before 0.00 0.50 mol 0.50mol Before 0.00 0.05 mol 0.05 mol

Addition Addition 0.01 mol — — 0.01 mol — —

After 0.00 mol 0.49 mol .51 mol After 0.00 mol 0.04 mol .06 mol

pH =pKa + log ( [base] / [acid] ) pH =pKa + log ( [base] / [acid] ) =5.00 + log ( 0.51 / 0.49 ) =5.00 + log ( 0.06 / 0.04 ) =5.02 =5.18 % change=(5.02 - 5.00) / 5.00 X100% % change=(5.18 - 5.00) / 5.00 X100% =0.4% =3.6% Buffer Range

Because a buffer should not differ by more than a factor of 10, we can use the Henderson-Hasselbalch equation to find a pH range:

pH = pK + log ( [base] / [acid] ) pH = pKa + log ( [base] / [acid] ) a = pK + log 10 = pKa + log 0.10 a = pK + 1 = pKa - 1 a ⇒ the effective pH range for a buffer solution is pKa ± 1 When making a buffer solution, use the pKa then adjust acid-conjugate base ratio Practice

A: Which acid would one use (combined with its sodium salt) to make a solution buffered at a pH of 4.25?

a.) HClO2 pKa = 1.95 b.) HNO2 pKa = 3.34 c.) HCHO2 pKa = 3.75 d.)HClO pKa = 7.54 Practice

B: Calculate the ratio of the conjugate base to the acid in formic acid (HCHO2) required to attain the desired pH of 4.25. (pKa = 3.74) pH = pKa + log ( [base] / [acid] ) 4.25 = 3.74 + log ( [base] / [acid] )

4.25 - 3.74 = log ( [base] / [acid] ) 0.51 = log ( [base] / [acid] )

[base] / [acid] = 100.51 = 3.24 Buffer Capacity

The amount of acid or base a buffer solution can neutralize without a significant change in pH

Buffer capacity increases with increasing absolute concentration of the buffer solution (increasing concentration of components)

Buffer capacity also increases as the concentration of acid and conjugate base reach similar levels (the [base] / [acid] approaches one)

However, if a buffer neutralizes mainly acid or mainly base it may have a much higher concentration of one of the components Titrations

Acid-Base titrations: an acidic (or basic) solution of unknown concentration reacts with a basic (or acidic) solution of known concentration in order to determine the original solution’s concentration Titrating

A pH indicator is mixed with the solution to monitor pH The solution of known acid or base concentration is slowly added in using a buret The acid and base neutralize each other as the titration continues When the equivalence point is reached the titration is complete (pH indicator- dye whose color depends on pH)

(Equivalence point- point when the number of moles of acid and base are stoichiometrically equal) Buret Strong Acid-Strong Base Titration

Ex: Titration of 25 mL 0.100M HCl with 0.100M NaOH Calculating equivalence point: ⇒ HCl (aq) + NaOH (aq) H2O (l) + NaCl (aq) Initial mol HCl = 0.025L x (0.100mol / 1 L) = 0.0025mol HCl NaOH solution = 0.0025mol x (1L / 0.100mol) = 0.0250L

The pH at the equivalence point will always be 7 (all + - H3O and OH ions have neutralized each other) To find the concentration of the initial solution convert the volume added to moles and then moles to concentration MaVa = MbVb

(mol acid) x (volume acid) = (mol base) x (volume acid)

( mol / L ) x (L) = ( mol / L ) x (L)

This equation is used to find the equivalence point Strong Acid-Strong Base Titration Continued + Initial pH= -log [H3O ] = -log (0.100) = 1 pH at any given point before equivalence point: mol NaOH added = (L added) x ( [NaOH] / 1 L ) + Because we are working with a strong acid and base H3O = mol HCl - mol NaOH + + [H3O ] = (mol H3O / total volume) + pH = -log [H3O ] pH at any given point after equivalence point: - - + ⇒ + -14 - ⇒ [OH ] = (mol OH - mol H3O ) / total volume [H3O ] = 10 / [OH ] pH = -log + [H3O ] Convert mL ⇒ mol, Use addition table to find mol H O+ Practice 3 + ⇒ + + ⇒ mol H3O [H3O ], [H3O ] pH

Calculate the pH after adding 5 mL of 0.100 M NaOH to 25 mL of 0.100 M HCl

- + ⇒ OH + H3O 2 H2O

Before Addition 0.00 0.0025 mol

Addition 0.0005 mol —

After Addition 0.00 mol 0.0020 mol

+ + [H3O ] = ( .0020 mol H3O ) / ( 0.0250 L + 0.00500 L ) = 0.0667 pH = -log 0.0667 = 1.18 Total Volume Titration curve/pH curve Strong Base-Strong Acid Titration College Board- Buffer College Board- Titrations