Titrations
Unit:  Acids & Bases
Skills:
•	calculate amounts of acid or base neutralized and pH in a titration
titration:  intentionally neutralizing an acid or base
Strong Acid/Strong Base
When a strong acid is titrated with a strong base, the net ionic reaction is:
H+ (aq) + OH− (aq) → H2O (ℓ)
Titration Curve
With a strong acid and strong base, water effectively acts as a buffer, and the equivalence point is at exactly pH 7.  A plot of pH vs. the amount (volume) of base added looks like the following:

Calculations:
To find the amount of acid or base required to neutralize a solution, treat the problem as a dilution (molarity) problem.  The total millimoles of H+ or OH− present equals the millimoles of acid or base needed.
Q:	How many mℓ of 0.100	M Ba(OH)2 are needed to neutralize 25 mℓ of 0.075 M HCl?
A:	
To find the pH of a solution, simply find [H+] as you would with any dilution (molarity) problem.  (However, remember that in pure water, 1.0 × 10−7 M of the [H+] comes from the natural ionization of H2O.  E.g., a solution of 10−12 M HCl would have a pH of 7, not 12.)

Q:	If 50. mℓ of 0.200 M HCl is titrated with 75 mℓ of 0.100 M NaOH, what is the final pH?
A:	
The 7.5 mmol of OH− neutralizes 7.5 mmol of the H+, which leaves 2.5 mmol H+ (= 0.0025 mol) still in solution.
The volume of the solution is 50. + 75 = 125 mℓ, so the concentration is:

Weak Acid/Strong Base
When a weak acid is titrated with a strong base, the weak acid and its conjugate base act as a buffer.  (The same is true for titration of a weak base with a strong acid.)  The titration curve looks like the following:

Note that the equivalence point is reached after the buffering capacity of the weak acid is exhausted.  Note also that the equivalency point is above 7.  (Similarly, for titration of a weak base by a strong acid, the final pH will be below 7.)
The calculation for the amount of base needed to titrate a weak acid (or vice-versa) is the same as for a strong acid (or base)—calculate the moles of H+ in the acid, which equals the moles of OH− needed.
pH calculations for weak acids require the Henderson-Hasselbalch equation.  For our sample problem, we will consider the titration of 10 mℓ of 0.5 M acetic acid using 0.5 M NaOH.  The Ka of acetic acid is 1.75 × 10−5.
The pH calculations are different in different regions of the titration curve:  before addition of any base, in the buffering region, at the equivalence point, and beyond the equivalence point.  We will consider each of these calculations separately.
Before Any Base is Added
This is simply an acid dissociation calculation, so we can calculate [H3O]+ directly from the formula for Ka:

Because 1 of each comes from dissociation of HAc, [H3O+] = [Ac−]:

Titrating In the Buffering Region
In the buffering region (the moles of base added are less than the moles of acid present), you need to do an I.C.E. calculation and use the Henderson-Hasselbalch equation. (Ignoring water as it is a pure liquid, not included in equilibrium expressions)
HAc (aq) + OH− (aq) → H2O (ℓ) + Ac− (aq)
	HAc	OH−	Ac−	
initial	5.0 mmol	x mmol	0	
change	−x mmol	−x mmol	+x mmol	
equilibrium	5.0 − x	0	x mmol	
Therefore, after adding base:
, , and 
From the Henderson-Hasselbalch equation:

E.g., if we had added 5.0 mℓ of base (2.5 mmol), this would work out to:

At the Equivalence Point
At the equivalence point, all of the weak acid (HAc) and strong base (OH−) have been neutralized, and only the conjugate base of the weak acid (Ac−) is present.  This means we now have a weak base dissolved and partially dissociated in water:
Ac− (aq) + H2O (ℓ)  HAc (aq) + OH− (aq)
Recall that, for the dissociation of a weak base, 
Recall also that Kw = Ka·Kb, which gives:

We can calculate [Ac−] from the amount of base added:

From the stoichiometry of the dissociation, we can substitute [HAc] = [OH−].  Therefore:

Beyond the Equivalence Point
Beyond the equivalence point, all of the acid has been neutralized and we have excess OH−.  At this point, the weak base Ac− is no longer a factor, and we calculate the pH from the pOH, which we calculate from [OH−].
E.g., if we had added a total of 11 mℓ (= 5.5 mmol) of NaOH, we would have used 5.0 mmol to neutralize the HAc, and we would have 0.5 mmol OH− remaining in solution.  The total volume is 10 + 11 = 21 mℓ.

pOH = −log(0.024) = 1.62
pH = 14 − pOH = 14 − 1.62 = 12.38

Polyprotic acids 

Overview

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