16.7: Acid-Base Titration Curves

Acid-Base Titrations:

An acid-base titration is a procedure for determining the amount of acid (or base) in a solution by determining the volume of base (or acid) of known concentration that will completely react with it.

Can titrate:
1) A strong acid with a strong base
2) A strong acid with a weak base
3) A weak acid with a strong base
4) A weak acid with a weak base → this kind of titration will not be on the exam.

An acid-base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid). Here is one for a strong acid titrated with a strong base:

The change in pH as the base is added will be slow at first, then it will reach a point where the pH increases drastically before leveling out again.

The middle of this rapid incline is called the equivalence point: the point in a titration when the smallest amount of titrant has been added that is sufficient to fully neutralize or react with the acid or base (the point when all the acid and base have reacted).

Example:

The case of a strong acid and a strong base:

Note: The pH at the equivalence point for a strong acid and a strong base will always be 7.

What is the pH when 27.0mL of 0.450 M HCl is titrated (read as mixed!) with 45.0mL of 0.500 M NaOH?

Remember with mixing/titrating, the first thing we do is convert to moles:

mols HCl = (0.0270 L)(0.450 mols/L) = 0.0122 mol HCl
mols NaOH = (0.0450 L)(0.500 mols/L) = 0.0225 mol NaOH

The acid and base react to form a salt:

HCl (aq) + NaOH (aq) → H₂O (l) + NaCl (aq)



Convert to concentration by dividing by new volume:

NaOH = (0.0103 mol) / (0.072 L) = 0.143 M NaOH = 0.143 M OH^-
pOH = -log(0.143) = 0.844
pH = 14 – pOH = 13.2

Example:

The case of a strong acid and a weak base:

What is the pH at the equivalence point when 0.150 M of HCl titrates 20 mL of 0.200 M NH₃?

Remember that at the equivalence point, mols of HCl = mols of NH₃.

Since we're dealing with a titration, first convert to mols:

mols NH₃= (0.200 mols/L)(0.0200 L) = 4.00 x 10^-3 mols NH₃
and mols NH₃ = mols HCl = 4.00 x 10^-3 mols

To get the final volume after titration, we'll need to know the volume of HCl. Since we know the mols and molarity, we can use what we know to find the volume.

Concentration = Mols x Volume,
Volume = Mols / Concentration

Volume HCl = (4.00 x 10^-3 mols) / (0.150 M) = 0.0267 L HCl
So total volume of titration is 0.0267L + 0.0200 L = 0.0467 L
Concentrations of HCl and NH₃ = (4.00 x 10^-3 mols) / (0.0467 L) = 0.0857 M

So 0.0857 M of the acid and 0.0857 M of the base react to form 0.0857 M of a salt:

HCl + NH₃ → NH₄Cl

NH₄Cl dissociates in water and leaves NH₄⁺ and Cl^- ions. Cl^- ions form a strong acid which dissociates 100% in water, so no reaction there. When NH₄⁺ ions react with water:

NH₄⁺ (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + NH₃ (aq)



K_a = (x²) / (0.0857 – x) → approximation holds
K_a= (x²) / (0.0857)

We need to find the K_a for NH₄⁺, so first find K_b of the conjugate base, NH₃, and plug it in:

K_a = K_w / K_b

K_b for NH₃ = 1.8 x 10^-5
K_a for NH₄⁺= (1 x 10^-14) / (1.8 x 10^-5) = 5.56 x 10^-10

5.56 x 10^-10 = (x²) / (0.0857)
x = 6.9 x 10^-6 = [H₃O⁺]
pH = -log(6.9 x 10^-6) = 5.16

Example:

The case of a weak acid and a strong base:

What is the pH when 35.0mL of 0.215 M HClO is titrated by 0.400 M LiOH to the equivalence point?

Remember that at equivalence, mols HClO = mols LiOH.

And since we're titrating, remember the first thing to do is convert to mols:

mols HClO = (0.350 L)(0.215 mols/L) = 0.00753 mols HClO
and mols HClO = mols LiOH = 0.00753 mols

To get the final volume after titration, we'll need to know the volume of LiOH. Since we know the mols and molarity, we can use what we know to find the volume.

Concentration = Mols x Volume,
Volume = Mols / Concentration

Volume LiOH = ( 0.00753 mols) / (0.400 M) = 0.0188 L LiOH
So total volume of titration is 0.0350L + 0.0188 L = 0.0538 L
Concentrations of HClO and LiOH = (0.00753 mols) / (0.0538 L) = 0.140 M

So 0.140 M of the acid and 0.140 M of the base react to form 0.140 M of a salt:

HClO + LiOH → LiClO

LiClO dissociates in water and leaves Li⁺ and ClO^- ions. Li⁺ ions form a strong base which dissociates 100% in water, so no reaction there. When ClO^- ions react with water:

ClO^- (aq) + H₂O (l) ⇌ HClO (aq) + OH^- (aq)



K_b = (x²) / (0.140 – x) → approximation holds
K_b = (x²) / (0.140)

We need to find the K_b for ClO^-, so first find K_a of the acid, HClO, and plug it in:

K_b = K_w / K_a

K_a for HClO = 3.5 x 10^-8
K_b for ClO^- = (1 x 10^-14) / (3.5 x 10^-8) = 2.86 x 10^-7

2.86 x 10^-7 = (x²) / (0.140)
x = 2.00 x 10^-4 = [OH^-]
pOH = -log(2.00 x 10^-4) = 3.7
pH = 14 – 3.7 = 10.3