# Formula for a titration curve?

• Nikitin
In summary, it is possible to find the second derivative of a titration curve on wikipedia. This curve is represented by the pH and in turn the pH is given by the -log[H+]
Nikitin
They say it's possible to find the growth-rate ie the second derivative of a titration curve on wikipedia. I am interested in finding & deriving such a formula, because I need to know why the growth-rate is so small in the beginning of the titration but it gets so much bigger as you add larger volumes of the titrant.. Shouldn't it be the opposite, if it isn't linear?

I don't get it, if a strong acid is titrated gradually by a strong base, shouldn't the pH growth-rate until the equivalence point be constant? Ie the function should be linear until the equivalence point?

If I would guess the reason, would it be because the water is kind of "resistant" to titration? Let's take the picture below as an example: there are kind of "reserves" of un-protolyzed acids which act as a buffer vs the titrating base?[PLAIN]http://www.files.chem.vt.edu/chem-ed/titration/graphics/titration-strong-acid-35ml.gif

Last edited by a moderator:
Nikitin said:
They say it's possible to find the growth-rate ie the second derivative of a titration curve on wikipedia. I am interested in finding & deriving such a formula, because I need to know why the growth-rate is so small in the beginning of the titration but it gets so much bigger as you add larger volumes of the titrant.. Shouldn't it be the opposite, if it isn't linear?

I don't get it, if a strong acid is titrated gradually by a strong base, shouldn't the pH growth-rate until the equivalence point be constant? Ie the function should be linear until the equivalence point?

If I would guess the reason, would it be because the water is kind of "resistant" to titration? Let's take the picture below as an example: there are kind of "reserves" of un-protolyzed acids which act as a buffer vs the titrating base?

[PLAIN]http://www.files.chem.vt.edu/chem-ed/titration/graphics/titration-strong-acid-35ml.gif[/QUOTE]

The curve represents the pH and in turn the pH is given by the -log[H+]

So you would expect the curve to follow a logarithmic curve as the hydrogen ion concentration falls by an even amount.

BUT the hydrogen ion concentration does not fall by an even amount as the volume of the mixture is constantly increasing while the molar amount of hydrogen is constantly decreasing.

The result of the three factors above is the complex curve shown. The best thing to do is calculate a few examples:

For the addition of NaOH (0.1M) to 25 ml HCl (0.1M)

Initially there is only 0.1M HCl so the pH = 1

Add 1ml of NaOH
-----------------

Moles of NaOH added = 0.001 x 0.1 = 0.0001
Moles of Hydrogen ions used up = 0.0001
Moles of H+ remaining = (0.025 x 0.1) - 0.0001 = 0.0024
New volume = 0.026 ml
New [H+] = 0.0024/0.026 = 0.0923
pH = 1.035

So, addition of 1 ml increases the pH by only 0.035 units

Add another 9 ml NaOH (total 10 ml)
----------------------------------

Moles of NaOH added = 0.01 x 0.1 = 0.001
Moles of Hydrogen ions used up = 0.001
Moles of H+ remaining = (0.025 x 0.1) - 0.001 = 0.0015
New volume = 0.035 ml
New [H+] = 0.0015/0.035 = 0.0429
pH = 1.37

So, addition of 10 ml total NaOH increases the pH by only 0.37 units

Add another 10 ml NaOH (total 20 ml)
----------------------------------

Moles of NaOH added = 0.02 x 0.1 = 0.002
Moles of Hydrogen ions used up = 0.002
Moles of H+ remaining = (0.025 x 0.1) - 0.002 = 0.0005
New volume = 0.045 ml
New [H+] = 0.0005/0.045 = 0.0111
pH = 1.95

So, addition of 20 ml total NaOH increases the pH by only 0.95 units

Add another 4 ml NaOH (total 24 ml)
----------------------------------

Moles of NaOH added = 0.024 x 0.1 = 0.0024
Moles of Hydrogen ions used up = 0.0024
Moles of H+ remaining = (0.025 x 0.1) - 0.0024 = 0.0001
New volume = 0.049 ml
New [H+] = 0.0001/0.049 = 0.00204
pH = 2.69

So, addition of 24 ml total NaOH increases the pH by only 1.69 units

But notice that the last 4 ml increased the pH by more than the prevous 10 ml

Add another 0.9 ml NaOH (total 24.9 ml)
----------------------------------

Moles of NaOH added = 0.0249 x 0.1 = 0.00249
Moles of Hydrogen ions used up = 0.00249
Moles of H+ remaining = (0.025 x 0.1) - 0.00249 = 0.00001
New volume = 0.0499 ml
New [H+] = 0.00001/0.0499 = 0.0002
pH = 3.70

So, addition of 24.9 ml total NaOH increases the pH by 2.7 units

But, the last 0.9ml changed the pH by over 2 pH units.

Now at neutralisation the pH is 7 so an addition of a further 0.1ml is going to change the pH from 3.7 to 7.0.

After the neutralisation point the pH is calculated by finding the excess [OH-] and using kw to get pH.

The result is that the curve is almost symmetrical, but not quite so as the volume keeps increasing.

Last edited by a moderator:
thanks, I get it now =)

Awesome, thanks so much for this explanation. I've been looking for a good explanation of the shape of a titration curve forever.

## 1. What is a titration curve?

A titration curve is a graphical representation of the pH of a solution as a function of the volume of a titrant added. It is used to determine the equivalence point and the concentration of an unknown solution.

## 2. What is the formula for a titration curve?

The formula for a titration curve is pH = pKa + log ([A-]/[HA]) where pH is the logarithm of the hydrogen ion concentration, pKa is the acid dissociation constant, and [A-] and [HA] are the concentrations of the conjugate base and the acid, respectively.

## 3. How do you plot a titration curve?

To plot a titration curve, you need to record the initial pH of the solution and then add small increments of the titrant while measuring the pH after each addition. The volume of titrant is plotted on the x-axis and the pH on the y-axis. Once the equivalence point is reached, the pH will rapidly change and then level off at a constant value.

## 4. What is the significance of the equivalence point on a titration curve?

The equivalence point is the point where the moles of the titrant added are equal to the moles of the analyte present. It is important because it indicates the completion of the reaction and allows for the determination of the concentration of the unknown solution.

## 5. How can a titration curve be used to determine the concentration of an unknown solution?

By measuring the volume of titrant at the equivalence point and using the known concentration of the titrant, the concentration of the unknown solution can be calculated using the formula M1V1 = M2V2. Additionally, the steeper the slope of the titration curve at the equivalence point, the more concentrated the unknown solution is.

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