Titration Calculations

An example of titration problem:

I have a 25.00 mL sample of a strong acid at an unknown concentration. After adding 13.62 mL of a 0.096 M NaOH solution, equivalence was reached. Determine the pH of the acid.

What is the goal of a titration problem?

What is significant about equivalence and why is that important?

Equivalence

At equivalence:Moles of H+ = Moles of OH- added

If we know the [OH-] added, then at equivalence it will be equal to the [H+] of the original acid.

…After addition of 13.62 mL of a 0.096 M NaOH solution, equivalence was reached.

How can we get the [OH-]?

Be part of the solution, not the problem!

13.62 mL NaOH 1 L = 0.01362 L NaOH added 1000 mL

0.096 M NaOH = X moles NaOH = 1.308 x10-3 mol OH-

0.01362 L NaOH

What does that number tell us?

1.308 x 10-3 mol OH- added = 1.308 x 10-3 mol H+ in original acid

An example of titration.

We calculated that there were 1.308 x 10-3 moles H+ in original sample. Are we done?

We need the pH of the acid. How do we do that?

pH = - log [H+] – We need the Molarity of H+

Molarity = 1.308x10-3mol / 0.025L= 0.052 M

pH = - log [0.052] = pH 1.28 (2 Dec. Places)

Problem: I have a 25.00 mL sample of a strong acid at an unknown concentration. After adding 13.62 mL of a

0.096 M NaOH solution, equivalence was reached. Determine the pH of the acid.

Titrations not to Equilibrium

Titrations are not required to go all the way until the equilibrium point.

Some problems ask how the pH of a solution will change as titration occurs.

Problem type #2

Determine the pH when 49.0 mL of NaOH is added to a 50.0 mL HCl solution.

Notice that equivalence or complete neutralization is not mentioned, nor obtained.

In these types of problems, we need to determine how many H+ ions and OH- ions are neutralized and then determine what ions are left over to affect the pH of the resulting solution.

Start with the Equation

NaOH + HCl H2O + NaCl

If we remove the aqueous spectator ions, the equation looks like this:

OH- + H+ H2O To solve this question, we need to know how

many of the ions are used to make the water.

Need to determine moles not M

Another tricky part about this question is that the volume will change when the two solutions combine.

When volume changes, Molarity changes so we need to convert everything to MOLES:

Molarity = Moles/Liters

Converting to Moles

NaOH = 0.100 M = X mol 4.90 x10-3 mol

0.049 L

HCl = 0.100 M = X mol 5.00 x10-3 mol

0.050 L

Repurposed ICE Chart

We need to calculate the concentrations at equilibrium. What did we have initially? What did we Change? and what is left over?

OH- + H+ H2O

Initial

Change

Equilibrium

What did we begin with?

We started with the hydrochloric acid. How much did we have?

Where will that go in the ICE Box?

OH- + H+ H2O Initial: 0 5.00x10-3

0

What did we change?

We added the sodium hydroxide.Where will that go in the ICE Box?How much did we add?

OH- + H+ H2O

Change 4.90x10-3 0

0

What did we end up with?

How do we determine the concentrations at equilibrium. (Initial + Change = Equilibrium)

But remember a neutralization reaction is taking place so we must make water.

Initial 0 5.00x10-3 0

Change 4.90x10-3 0 0

Eq. 4.90x10-3 5.00x10-3 0

OH- + H+ H2O

What is left over at equilibrium?

The question remains, how much water is made? In this reaction all possible reactants are used up.

Equilibrium 4.90x10-3 5.00x10-3 4.90x10-3

OH- + H+ H2O

So we are limited by the amount of the [OH-] ion. That means we have excess [H+] ions that will affect the pH of the solution!!

How many ions are left over?

So how many moles of [H+] do not react? 5.00 x 10-3 - 4.90 x 10-3 = 1.0 x 10-4 mol H+

These ions remain in the solution and will affect the pH of the solution.

So we have the moles of H+, but what do we need to determine the pH?

Eq. 4.90x10-3 5.00x10-3 4.90x10-3

OH- + H+ H2O

Tricky Volume!

We need the Molarity! But this is tricky because the volume of the solution changed when we added the acid to the base. (Molarity = Moles / Liters)

What is the final volume of the solution?

49.0 mL of NaOH + 50.0 mL of HCl = 99.0 mL

1.0 x 10-4 / 0.099 L = 1.01 x10-4 [H+]

And Finally the pH part!

pH = -log (1.0 x10-3 [H+]) pH = 3.00 This makes sense (hopefully)

because we’re not at equilibrium so the pH isn’t at 7.