Titration Problem Solving Guide by Samuel Zapantis

How to solve a Titration problem (Weak-Strong)

samuelzapantis — 2024-04-21 00:11:17 UTC

This is your guide to dealing and solving those pesky titration problems

Beginning of Titration

samuelzapantis — 2024-04-21 00:23:42 UTC

To begin, our titration problem is as follows; Find the pH of the following increments for a titration of 25 mL of 0.3 M HF with 0.3 M NaOH, the ka value is 6.6x10^-4:

Initial pH
After adding 10 mL of 0.3 M NaOH.
After adding 25 mL of 0.3 M NaOH.
After adding 30 mL of 0.3 M NaOH.

Initial pH

samuelzapantis — 2024-04-21 00:32:38 UTC

To calculate initial pH, an ICE table must be used to acquire [H⁺]. Once we have acquired the [H⁺], then you can take the negative log of it and get the initial pH.

See work on next tab

Set up the dissociation reaction of HF at equilibrium.
After setting up the chemical equation begin filling out the ice table with the information you are given. The concentration of HF is known to be 0.3 M; but not for H⁺ or F^-, so those will be filled with x's.
Once the ice table has been filled out with known concentration and the variables for the unknown, begin solving for the x's. To make the solving easier, if the Ka has a difference of larger than 10³ compared to the concentration of HF, the "-x" can be canceled out.
The Ka ( Acid Dissociation Constant) of HF is given and is 6.6 x 10^-4. This will be used in the equilibrium expression to solve for "x". To set up your equilibrium expression it must be in the form of Ka=products/ reactants.
With your equation set up all that needs to be done now is solving for x by isolating. Once the isolation of x is done you will be left with the concentration of [H₃O⁺]/[H⁺].
Since pH is directly related to [H⁺], to acquire the initial pH; negative log the [H⁺]. The outcome of that will be your initial pH.
Finally, the initial pH of the solution is 1.85

Initial pH ( worksheet)

samuelzapantis — 2024-04-21 00:39:29 UTC

pH after adding 10 mL of 0.3 M NaOH (Buffer)

samuelzapantis — 2024-04-21 04:18:15 UTC

After acquiring the initial pH of HF, now we move on to the real stuff! Before we hop into it, first the most important thing you must calculate before any titration problem is the equivalence volume. Calculating this will tell you at what volume the equivalence point is, the rearranged formula is as follows: M₁V₁/M₂=V₂; V₂is equal to your volume at equivalence.

See work on next tab

Write down the acid-base neutralization reaction between HF and NaOH.
Using the molarities and volumes that are provided, you want to calculate moles by multiplying molarity by volume in liters.
Now everyone has different ways of going about titration problems, but I aim to make it as simple as can be. So, once the number of moles of each compound is calculated simply just subtract them from each other.
Once the difference is calculated you will now have the number of moles of conjugate base that is produced as well as, the number of moles of acid left after titration. 0.0045 moles of HF will be left over and 0.003 moles of NaF or F^- will be produced from the NaOH. This part which is before equivalence is the buffer piece of the titration.
Now that the moles of Conjugate base and acid are calculated, we can move on to using the Henderson-Hasselbalch equation for buffers. The Henderson-Hasselbalch equation is as follows :
pH=pka + log([Conjugate base]/[Acid])
- pKa is simply just the negative log of the Ka
- Although it asks for the concentration of both C.B. and Acid; due to them having the same total volume it will result in the volumes canceling out. Leaving you to just use the moles to calculate with instead.
Finally, after inputting all the calculated information, the pH after adding 10 mL of 0.3 M NaOH is 3.00.

pH after adding 10 mL of 0.3 M NaOH (Buffer) Worksheet

samuelzapantis — 2024-04-21 04:27:12 UTC

pH after adding 25 mL of 0.3 M NaOH (Equivalence)

samuelzapantis — 2024-04-21 15:46:34 UTC

Now we are getting to the juicy part! One thing to note before jumping in, at equivalence moles of acid=moles of base meaning both will be used up only leaving the conjugate base that was created.

See work on next tab

Begin by creating a BCA table ( Before , Change, and After). This allows you to calculate the change in moles as reactants are used up and products are formed.
As the moles are always the same all the reactants will be used up leaving just the conjugate base (or conjugate acid depending on the nature of the problem).
After calculations, you will be left with 0.0075 moles of F^-, and a total volume of solution of 50 mL (the volume of both acid and base are added together).
To continue, using the calculated moles and total volume, calculate the molarity of F^-; the outcome should be 0.15 M F^-
Now you have the concentration of conjugate base and no remaining reactants as the titration has reached equivalence. To calculate the pH we must do yet another... ICE table.
In addition, before doing the ice table you want to begin by flipping the initial chemical equation. Instead of F^-as the product it now becomes the main reactant, and OH^- & HF become the products.
After setting up the ICE table you then begin to fill out any known values and the unknown "x's". For this ICE table the concentration of OH^- will be solved for meaning you must know your conversions!
Before you solve for the "x", you must convert Ka to Kb as the K you have now is for acids, but this is a base dissociation reaction as it has been flipped. To acquire the Kb , simply just divide the Kw ( ion product constant) by the Ka (acid dissociation constant).
Once the Kb is found, it is now possible to solve for the "x" or [OH^-]. In addition due to the Kb having a difference of larger than 10³ compared to the concentration of F^-, the "-x" can be canceled out. This makes it easier to solve for the unknown.
After careful calculations, the "x" should come out to 1.51x10^-6, but when this is negative logged it will give you the pOH. No worries, once you acquire the pOH just subtract 14-pOH and you will be given the pH of the solution at equivalence.
To conclude, the pH at equivalence should come out to 8.18.

pH after adding 25 mL of 0.3 M NaOH ( Equivalence) Worksheet

samuelzapantis — 2024-04-21 15:51:41 UTC

pH after adding 30 mL of 0.3 M NaOH (Excess)

samuelzapantis — 2024-04-21 16:21:34 UTC

This is the easiest of them all as the work becomes minimal and does not take much time. This part deals with when there is excess titrant in the solution.

Gather the already calculated moles of HF, and calculate the new moles of NaOH when the volume is 30 mL. After multiplying molarity by volume you will be left with 0.009 moles of NaOH, this is more moles of base than acid.
To get the number of moles left in excess simply just subtract moles of NaOH by moles of HF, this will give you an outcome of 0.0015 moles NaOH (excess base). After getting moles of excess, divide the moles by the total volume added to the solution, in this case, it would be 55 mL ( 30 mL + 25 mL). The division of these two parts will offer the [NaOH].
Once you find the [NaOH], which should be 0.027 M. Then look at the base you are dealing with, for strong bases such as NaOH; the compound completely dissociates when in solution.
With a complete dissociation, the molarity of the initial compounds is the same for its separate molecules. So, the dissociation will result in gaining Na⁺& OH^-, and they will have the same concentration as the one calculated for NaOH.
In addition, notice how we have an OH^- molecule, this can be used to gain the pH similar to how we did in the previous problem. Take the negative log of [OH^-], this will net you a pOH of 1.57; finally, just subtract 14 - pOH and you will have the pH of the excess solution.
The pH of the excess problem comes out to 12.43.

pH after adding 30 mL of 0.3 M NaOH (Excess) Worksheet

samuelzapantis — 2024-04-21 16:25:09 UTC

Congratulations! You now know how to do a full titration problem.

samuelzapantis — 2024-04-21 16:31:22 UTC

Now go and practice till those fingers fall off!