How Do You Calculate pH Changes in a Phosphoric Acid Titration?

[lananh]

Homework Statement

I've been working on this problem and am having issues with some of the problems...

all we learned was that EQ is pH= pka and at eq and EQ1/2 is pk1+pk2/2 which can't even be applied to this question due to x values.

I would GREATLY appreciate ANY help/guidance/etc.

Calculate the pH of 100.0 mL of a 0.110 M phosphoric acid solution being titrated with 0.110 M KOH under the following conditions:

pH = when no KOH has been added. SOLVED 1.598

pH = 2.26 when 50.0 mL of KOH has been added. Hint: Ka1 is not small. (*) SOLVED

pH = when 100.0 mL of KOH has been added.(*)

pH = when 150.0 mL of KOH has been added.

pH = when 200.0 mL of KOH has been added.(*)

pH = when 250.0 mL of KOH has been added. Hint: Ka3 is very small. (*)

pH = 12.29 when 300.0 mL of KOH has been added. SOLVED

pH = when 400.0 mL of KOH has been added.

but all the other 1/2 and eq points I am having issues with.. help please!

Homework Equations

not ph= pka at EQ and not ph= pka+pka2/2 at 1/2eq

The Attempt at a Solution

(above)

 

[nate808]

I understand that you are having trouble with some of the calculations for the pH values in this problem. Let me provide some guidance that may help you.

Firstly, it is important to remember that in a titration, the pH of the solution is determined by the concentration of the acid and base, as well as the dissociation constants (pKa) of the acid. In this case, you are titrating a phosphoric acid solution with a strong base, KOH.

To solve for the pH at each point, you will need to use the Henderson-Hasselbalch equation, which is pH = pKa + log([base]/[acid]). However, since phosphoric acid is a polyprotic acid (meaning it can donate more than one proton), you will need to consider the dissociation constants for each proton.

For the first pH value, when no KOH has been added, the solution is simply the 0.110 M phosphoric acid solution. This means that the concentration of the acid is 0.110 M and the concentration of the base is 0 M. You can use this information to calculate the pH using the Henderson-Hasselbalch equation.

For the second pH value, when 50.0 mL of KOH has been added, you will need to first determine the moles of KOH added and subtract that from the initial moles of phosphoric acid. This will give you the new concentration of the acid and base, which you can then use in the Henderson-Hasselbalch equation to solve for the pH.

For the third pH value, when 100.0 mL of KOH has been added, you will need to consider the dissociation constants for both the first and second protons of phosphoric acid. This means that you will have two equations and two unknowns (pH and [base]/[acid]). You can solve these equations simultaneously to find the pH value.

For the fourth and fifth pH values, when 150.0 mL and 200.0 mL of KOH have been added, respectively, you will need to consider the dissociation constants for all three protons of phosphoric acid. Again, this means you will have three equations and three unknowns to solve for the pH value.

For the sixth pH value, when 250.0 mL of KOH has been added, you will need to consider the dissociation constants for the third proton

 

[Blueshift5]

Hello,

It seems like you are struggling with your titration calculations for phosphoric acid. Let's break down the problem and see if we can guide you towards the solution.

Firstly, it is important to note that phosphoric acid (H3PO4) is a polyprotic acid, meaning it can donate multiple protons in solution. In this case, it has three acidic protons, with corresponding acid dissociation constants (Ka) of Ka1 = 7.5 x 10^-3, Ka2 = 6.2 x 10^-8, and Ka3 = 4.8 x 10^-13.

Now, let's look at the given information and see how we can approach the problem. We are titrating 100.0 mL of a 0.110 M phosphoric acid solution with 0.110 M KOH. This means that at the start of the titration (when no KOH has been added), we have a solution of H3PO4 with a concentration of 0.110 M.

To find the pH at this point, we can use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base (H2PO4-) and [HA] is the concentration of the acid (H3PO4). Since we are at the start of the titration, all of the H3PO4 is still present and none of it has been protonated, so [A-] = 0 and [HA] = 0.110 M. Therefore, the pH = pKa1 = 2.26.

For the second part of the titration, when 50.0 mL of KOH has been added, we have a solution with a total volume of 150.0 mL and a concentration of 0.073 M (since 50.0 mL of the original 0.110 M solution has been neutralized). At this point, we can use the Henderson-Hasselbalch equation again, but this time with the Ka2 value, since the first proton has been deprotonated. This will give us a pH = pKa2 + log([H2PO4-]/[H3PO4]). You can use this same approach for the other points in the titration, using the appropriate Ka value for each step.

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