Titration of two acid by a strong base

[Betterave]

Hi, i REALLY stucked in this pb :

50 mL of hydrochloric acid 0.2 mol/L are added to 50 mL of propanoic acid of the same titer and pKa = 4.8. The mixture is titrated with NaOH with a concentration of 1.0 mol/L
Calculate the pH corresponding to the addition of: 9 mL; 10mL; 15 mL and 20 mL of NaOH

1) The reaction between HCl and NaOH is H+ + OH- —> H₂O
2) Then i’ve done an ice table and i found out that pH is 2 for 9ml

Buut this is wrong, tell me why

[ChenBeier]

How die you calculate?

[Betterave]

I have done an ice table :
H+ + OH- —> H₂O

Moles of H+ : n = C*V = 0.2*0.05 = 0.01 mol
Moles of OH- : n = C*V = 1*0.009= 0.009 mol
Moles of H₂O : exces

Then we calculate the reagent who will finish first : it’s OH-
So at final state we have 0.001 mol of H+ left

Calculate the resulting concentration : c=n/V = 0.001/0.1 = 0.01 mol/L

PH=-log(H+) = -log(0.01) = 2

Apparently it’s wrong so help me

[ChenBeier]

We have a Volumen of 50 ml HCl and 50 ml C₂H₅COOH and 9 ml NaOH total 109 ml
The molarity gives 10 mmol HCl, 10 mmol C₂H₅COOH and 9 mmol NaOH
The Neutralization gives 1 mmol HCl, 10 mmol C₂H₅COOH. HCl is the stronger acid so the propanoic acid is negligible. So we have 1 mmol in 109 ml gives 0.009 mol/l what gives a pH of 2,04.

[Betterave]

Thanks for your replies,
In the exercise they that the Volume of NaOH added can be neglected to facilitate the calculations - so i was close
But someone told me it’s wrong and we have to calculate the VE₁ and VE₂, and then make an ICE table

For 9ml of NaOH added the pH = 2.04

[ChenBeier]

For 9ml of NaOH added the pH = 2.04

That is what I calculated.

What is VE₁ and VE₂?

[Betterave]

Yes thanks for your replies and calculation

VE₁ = the volume at which the exact number of moles of NaOH are added to react with all HCl
VE₂ = the volume at which the exact number of moles of NaOH are added to react with all C₂H₅COOH

[ChenBeier]

Yes this will be important for the next volumes. In the first case is VE₁ = 9 ml = 9 mmol and VE₂ = 0 ml = 0 mmol.

[JustinPlaying]

The error in your initial calculation may have stemmed from not correctly accounting for the presence of propanoic acid and its equilibrium after neutralizing all of the HCl. Always consider the state of both acids and their dissociation when calculating pH in such titration scenarios.

[Benzene]

To begin, we need to determine the amounts of hydrochloric acid (HCl) and propanoic acid (CH₃CH₂COOH) in the mixture. Both HCl and CH₃CH₂COOH are at a concentration of 0.2 mol/L, and we're adding 50 mL of each solution. To find the number of moles, we multiply the concentration by the volume. For HCl, that’s 0.2 mol/L × 0.05 L = 0.01 mol. The same calculation applies to CH₃CH₂COOH: 0.2 mol/L × 0.05 L = 0.01 mol.

Since the pKa value for propanoic acid is 4.8, initially, the concentration of CH₃CH₂COOH^⁻ ions is much lower than the concentration of undissociated CH₃CH₂COOH molecules. Thus, we can assume that the starting pH is close to the pKa value of 4.8.

Next, let’s determine the amount of NaOH added in each step:

For 9 mL of NaOH, multiply the concentration (1.0 mol/L) by the volume (0.009 L) to get 0.009 mol.
For 10 mL of NaOH, the calculation is the same: 1.0 mol/L × 0.01 L = 0.01 mol.
For 15 mL of NaOH: 1.0 mol/L × 0.015 L = 0.015 mol.
For 20 mL of NaOH: 1.0 mol/L × 0.02 L = 0.02 mol.

I apologise for the mistakes.
Corrections:
- Since all of the HCl has been neutralised and there is no longer any buffering from propanoic acid, the pH will go to a very high pH after 10 mL of NaOH.
- We have passed the equivalence point after adding 15 or 20 millilitres of NaOH, and the pH will continue to rise towards the basic environment.
- Since extra OH^⁻ from NaOH, not unreacted acid, is involved, the method for computing negative mole values is illogical in this situation.
-The Henderson-Hasselbalch equation is only helpful up to the equivalency threshold, beyond which the extra strong base (NaOH) controls the pH.
Tiredness was playing a nasty trick on me.
@ChenBeier: Thank you for your valuable input.

[ChenBeier]

The last part is nonsense, after all acid is neutralized. You have still CH₃CH₂COO- Ions and not Zero. What is CH₃CH₂COOH^⁻ ion by the way? Also more NaOH will rise pH according it's amount.
There is nothing with negative number.
Do you know how a titration curve looks like?

[ChenBeier]

Correct explaned.

[sofiacals]

Can this whole process be summarized in chemical equations?
tap tap shots