Understanding Titration Curves: pH Changes during Acid-Base Reactions, Study Guides, Projects, Research of Stoichiometry

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The equivalence point of the titration is the point at which exactly enough titrant has been added to react with all of the substance being titrated with no titrant left over. In other words, at the equivalence point, the number of moles of titrant added so far corresponds exactly to the number of moles of substance being titrated according to the reaction stoichiometry. (In an acidbase titration, there is a 1:1 acid:base stoichiometry, so the equivalence point is the point where the moles of titrant added equals the moles of substance initially in the solution being titrated.) Here is an example of a titration curve, produced when a strong base is added to a strong acid. This curve shows how pH varies as 0.100 M NaOH is added to 50. mL of 0.100 M HCl.

Calculate the pH at any point, including the

equivalence point, in an acidbase titration.

At the equivalence point, the pH = 7.00 for strong acidstrong base titrations. However, in other types of titrations, this is not the case The original number of moles of H

in the solution is:

What is the pH when 49.00 mL of 0.100 M NaOH solution have been added to 50.00 mL of 0.100 M HCl solution? Because it is a strong acidbase reaction, the reaction simplifies to: H

(aq) + OH (aq) H 2 O (l) The original number of moles of H

in the solution is:

What is the pH when 49.00 mL of 0.100 M NaOH solution have been added to 50.00 mL of 0.100 M HCl solution? Because it is a strong acidbase reaction, the reaction simplifies to: H

(aq) + OH (aq) H 2 O (l) The original number of moles of H

in the solution is: 50.00 x 10 3 L x 0.100 M HCl = 5.00 x 10 3 moles The number of moles of OH added is : 49.00 x 10 3 L x 0.100 M OH = 4.90 x 10 3 moles

What is the pH when 49.00 mL of 0.100 M NaOH solution have been added to 50.00 mL of 0.100 M HCl solution? Because it is a strong acidbase reaction, the reaction simplifies to: H

(aq) + OH (aq) H 2 O (l) The original number of moles of H

in the solution is: 50.00 x 10 3 L x 0.100 M HCl = 5.00 x 10 3 moles The number of moles of OH added is : 49.00 x 10 3 L x 0.100 M OH = 4.90 x 10 3 moles Thus there remains: (5.00 x 10 3 ) (4.90 x 10 3 ) = 1.0 x 10 4 moles H

(aq) The total volume of solution is 0.04900 L + 0.05000 L = 0.09900 L [H

] = {1.0 x 10 4 moles / 0.09900 L } = 1.0 x 10 3 M pH = 3.

EXAMPLE: What is the pH when 30.0 mL of 0.100 M NaOH have been added to 50.0 mL of 0.100 M acetic acid? STEP 1: Stochiometric calculation: The original number of moles of HC 2 H 3 O 2 in the solution is : 50.0 x 10 3 L x 0.100 M = 5.00 x 10 3 moles HC 2 H 3 O 2 Similarly, there are 3.00 x 10 3 moles of OH due to the NaOH solution. The reaction goes to completion: OH (aq) + HC 2 H 3 O 2 (aq) > C 2 H 3 O 2 (aq) + H 2 O (l) The total volume is 80.0 mL. We now calculate the resulting molarities : [HC 2 H 3 O 2 ] = { 2.00 x 10 3 mol HC 2 H 3 O 2 / 0.0800 L } = 0.0250 M [C 2 H 3 O 2 ] = { 3.00 x 10 3 mol C 2 H 3 O 2 } / 0.0800 L } = 0.0375 M STEP 2: Equilibrium calculation, using simplification: Ka = { [H

][C 2 H 3 O 2 ] / [HC 2 H 3 O 2 ] } = 1.8 x 10 5 [H

] = { KA [HC 2 H 3 O 2 ] / [C 2 H 3 O 2 ] } = { (1.8 x 10 5 )(0.0250) / (0.0375) } = 1.2 x 10 5 M pH = log(1.2 x 10 5 ) = 4.

Here, 0.100 M HCl is being added to 50.0 mL of 0.100 M ammonia solution. As in the weak acidstrong base titration, there are three major differences between this curve (in blue) and a strong basestrong acid one (in black): (Note that the strong basestrong acid titration curve is identical to the strong acidstrong base titration, but flipped vertically.)

1. The weakacid solution has a lower initial pH.
2. The pH drops more rapidly at the start, but less rapidly near the equivalence point.
3. The pH at the equivalence point does not equal 7.00. POINT OF EMPHASIS : The equivalence point for a weak basestrong acid titration has a pH < 7.00. Titration curve of a weak base being titrated by a strong acid:

Uses of Titration Curves Use titration data or a titration curve to calculate reaction quantities such as the concentration of the substance being titrated.

The most common use of titrations is for measuring

unknown concentrations. This is done by titrating a known

volume of the unknown solution with a solution of known

concentration (where the two react in a predictable

manner) and finding the volume of titrant needed to reach

the equivalence point using some method appropriate to

the particular reaction. Then, the volume and

concentration of titrant can be used to calculate the moles

of titrant added, which, when used with the reaction

stoichiometry, gives the number of moles of substance

being titrated. Finally, this quantity, along with the volume

of substance being titrated, gives the unknown

concentration.

For acidbase titrations, the equivalence point can be

found very easily. A pH meter is simply placed in the

solution being titrated and the pH is measured after various

volumes of titrant have been added to produce a titration

curve. The equivalence point can then be read off the

curve.

EXAMPLE:

If 80.0 mL of 0.200 M NaOH are required to reach the equivalence point

in a titration of 50.0 mL of hydrofluoric acid, what is the concentration of

the hydrofluoric acid?

The neutralization reaction goes to completion:

HF + OH F + H 2 O

The number of moles of NaOH added was:

nNaOH = [NaOH] x VNaOH

nNaOH = 0.200 M x 80.0 mL

nNaOH = 16 mmol

Since each NaOH produces 1 OH , nOH = nNaOH = 16 mmol

From the 1:1 stoichiometry between HF and OH , nHF = nOH = 16 mmol

So, the concentration of the original hydrofluoric acid solution was:

[HF] = nHF / VHF

[HF] = 16 mmol / 50 mL

[HF] = 0.320 M

In the same way, knowing the equivalence point can also be used to

calculate other unknown quantities of interest in acid base reactions, such as

concentration of titrant or volume of solution being titrated, provided that

enough other information is known to perform the calculations.