A 25.0 mL sample of 0.100 M propanoic acid (HC3H5O2, Ka = 1.3 ✕ 10-5) is titrated with 0.100 M KOH solution. after the addition of the following amounts of KOH with sig figs.
a) 24.5 ml
b) 25.0 ml
c) 30.0 ml
I worked a problem similar to this earlier today. Here it is.
https://www.jiskha.com/subjects?subject=chemistry
The differences are this is an acid and pyridine is a base but that doesn't change the process at all. The equation of propanoic acid, if we call it HPr, is
HPr + KOH ==> KPr + H2O
KPr is a salt, potassium propionate
a is 24.5 mL that is handled the same way as my other post.
b is the equivalence point and handled the same way
c. is the same as the 25, 28, 30 mL in the other post.
For the pH of part b, the equation for the salt is
KPr + HOH ==> HPr + OH^- + K^+
Let me know if you run into problems or have questions.
Hi Dr.Bob, I can't see your other post for some reason, and I had a question about the same things and I still don't understand how to do it.
Oh, I'm so excited to talk about titration! It really brings out the thrill in chemistry! So, let's get started with your question.
a) After adding 24.5 mL of KOH, we don't have to worry about sig figs because the volume is the same as the initial volume of propanoic acid. This means we haven't reached the equivalence point yet. So, we're still in the land of undissociated propanoic acid, where everyone is just hanging out, not breaking up. How cozy!
b) Ah, 25.0 mL of KOH, the same volume as the propanoic acid. This is where things get interesting! We have reached the thrilling equivalence point. It's like a dance-off between the propanoic acid and KOH. Who doesn't love a good dance battle, right? Anyway, at this point, all the propanoic acid has reacted with the KOH. The moles of KOH added is equal to the moles of propanoic acid initially present. Such chemistry magic!
c) Ah, finally, 30.0 mL of KOH. Who invited the overachiever to the titration party? We are well past the equivalence point now, where we have added excess KOH. So, now, it's like throwing a massive cup of KOH into a small bowl of propanoic acid, causing it to overflow. Oh no! The propanoic acid is completely neutralized, and we're left with excess KOH.
I hope that helps explain what's going on in this titration adventure. Chemistry can be such a blast! Enjoy the experiment!
To solve this problem, we will use the concept of acid-base titration. The balanced equation for the reaction between propanoic acid (HC3H5O2) and potassium hydroxide (KOH) is:
HC3H5O2 + KOH -> KC3H5O2 + H2O
The stoichiometry of the reaction tells us that one mole of propanoic acid reacts with one mole of potassium hydroxide.
a) 24.5 mL of KOH solution is added:
To determine the amount of propanoic acid remaining after the addition of 24.5 mL of KOH, we need to calculate the number of moles of KOH reacted. Since the concentration of KOH is given as 0.100 M, we can use the formula:
Moles of KOH = Volume of KOH (L) x Concentration of KOH (M)
Moles of KOH = 0.0245 L x 0.100 M = 0.00245 mol
Since the stoichiometry tells us that one mole of propanoic acid reacts with one mole of KOH, this means that 0.00245 mol of propanoic acid has reacted.
To find the amount of propanoic acid remaining, we subtract the moles of KOH reacted from the initial number of moles of propanoic acid:
Moles of propanoic acid remaining = Initial moles of propanoic acid - Moles of KOH reacted
Since we have 0.0250 L (25.0 mL) of 0.100 M propanoic acid, the initial number of moles is:
Initial moles of propanoic acid = Volume of propanoic acid (L) x Concentration of propanoic acid (M)
Initial moles of propanoic acid = 0.0250 L x 0.100 M = 0.00250 mol
Moles of propanoic acid remaining = 0.00250 mol - 0.00245 mol = 0.00005 mol
b) 25.0 mL of KOH solution is added:
Using the same calculations as in part a), we find that the moles of propanoic acid remaining after the addition of 25.0 mL of KOH is also 0.00005 mol.
c) 30.0 mL of KOH solution is added:
Using the same calculations as in part a), we find that the moles of propanoic acid remaining after the addition of 30.0 mL of KOH is 0.00005 mol.
In summary:
a) 24.5 mL of KOH solution: The moles of propanoic acid remaining = 0.00005 mol
b) 25.0 mL of KOH solution: The moles of propanoic acid remaining = 0.00005 mol
c) 30.0 mL of KOH solution: The moles of propanoic acid remaining = 0.00005 mol
To determine the pH of a solution during a titration, we need to calculate the number of moles of acid and base at each point and then determine the concentrations of the acid and conjugate base.
In this case, we are titrating propanoic acid (HC3H5O2) with KOH (potassium hydroxide). The reaction between propanoic acid and KOH forms potassium propanoate (KC3H5O2) and water.
a) 24.5 mL KOH added:
To calculate the moles of KOH, we first need to convert the volume to liters:
24.5 mL = 0.0245 L
Since the KOH concentration is given as 0.100 M, we can calculate the moles of KOH added:
moles of KOH = volume (L) x concentration (M) = 0.0245 L x 0.100 M = 0.00245 moles
Since KOH and propanoic acid have a 1:1 stoichiometry, the number of moles of propanoic acid consumed is also 0.00245 moles.
Now, we need to calculate the remaining moles of propanoic acid in the solution:
initial moles of propanoic acid - moles of propanoic acid consumed
= (volume of propanoic acid x concentration of propanoic acid) - moles of propanoic acid consumed
Since the initial volume of propanoic acid is given as 25.0 mL, we need to first convert it to liters:
25.0 mL = 0.0250 L
Now, let's calculate the remaining moles of propanoic acid:
remaining moles of propanoic acid = (0.0250 L x 0.100 M) - 0.00245 moles = 0.00255 moles
To determine the concentration of the acid, divide the remaining moles of acid by the volume of the solution:
Acid concentration = remaining moles of acid / volume of solution = 0.00255 moles / 0.0250 L = 0.102 M
Now, we can calculate the pKa of propanoic acid using the expression pKa = -log(Ka):
pKa = -log(1.3 ✕ 10-5) = 4.89
To calculate the pH, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid.
In this case, since we are dealing with a neutralization reaction, the concentration of the base is equal to the concentration of the conjugate acid. Therefore, [A-] = [HA] = 0.102 M.
pH = 4.89 + log(0.102/0.102) = 4.89 + log(1) = 4.89
Therefore, when 24.5 mL of KOH is added to the propanoic acid solution, the pH is 4.89.
b) 25.0 mL KOH added:
When 25.0 mL of KOH is added, the reaction is at the equivalence point, where the moles of acid consumed are equal to the moles of base added. Therefore, all the propanoic acid has been neutralized by KOH, and we are left with only the conjugate base, potassium propanoate.
Since we have only the conjugate base and no acid, the pH will be determined by the hydrolysis of the conjugate base. The solution will be basic.
c) 30.0 mL KOH added:
When 30.0 mL of KOH is added, excess base is present. The pH will depend on the concentration of the excess base and the salt formed from the reaction.
To determine the pH, we would need additional information, such as the concentrations of potassium propanoate and the remaining KOH.