A 30.0 mL sample of 0.165 molL−1 propanoic acid (Ka=1.3×10−5) is titrated with 0.300 molL−1KOH.
Calculate the pH at 0, 5,10, at equivalence point, 1/2 of the equivalence point, 20, 25 mL of added base.
at 0mL base pH = 2.83
at 5mL base pH = 4.52
at 10mL base pH = 5.07
at equivalance point pH = 8.96
at half the equivalance point pH = 4.88
at 20mL base pH = 12.32
at 25mL base pH = 12.67
your welcome.
nothing
Six problems in here. What do you know about any of these?
Ah, titration! The chemist's version of a game show, where acids and bases go head-to-head for the title of "Neutralization Nation." Let's get ready to pH-rumble!
At the start (0 mL), we have propanoic acid. Since propanoic acid is a weak acid, we can use the helpful Ka value of 1.3x10^-5 to our advantage. But let's not get too cocky, or we'll end up like a clown in the boxing ring!
The first step is to calculate the initial amount of moles of propanoic acid in our 30.0 mL sample:
moles of propanoic acid = concentration x volume
moles of propanoic acid = (0.165 mol/L) x (0.030 L)
moles of propanoic acid = 0.00495 mol
Now, let's consider the neutralization reaction between propanoic acid (HC3H5O2) and KOH. The equation looks like this:
HC3H5O2 + KOH -> KC3H5O2 + H2O
When KOH is added, it will react with the propanoic acid, creating the propanoate ion (C3H5O2-) and water. The moles of KOH added will determine the pH of our solution.
Let's calculate the pH at different points:
1. At 0 mL of KOH added: No KOH has been added yet, so we're still dealing with only propanoic acid. We can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = pKa + log([C3H5O2-]/[HC3H5O2])
pH = -log(1.3x10^-5) + log(0/0.00495) (Since no KOH has been added, there is no propanoate ion yet)
pH = -(-4.89) + log(0) (Don't worry, I won't divide by zero, this is just a joke!)
pH = 4.89 (Or we can say, it's pretty acidic, like a lemonade with extra lemons!)
2. At 5 mL of KOH added: Now we have some propanoate ion formed, but we need to determine the concentration. We've added 5 mL of 0.300 mol/L KOH, so the moles of KOH added are:
moles of KOH added = concentration x volume
moles of KOH added = (0.300 mol/L) x (0.005 L)
moles of KOH added = 0.0015 mol
To determine the concentration of propanoate ion, we'll need to consider the stoichiometry of the reaction. For every mole of propanoic acid (HC3H5O2), one mole of propanoate ion (C3H5O2-) is formed. So, the concentration of propanoate ion at 5 mL is:
concentration of propanoate ion = moles of propanoate ion / total volume
concentration of propanoate ion = 0.0015 mol / (0.030 L + 0.005 L)
concentration of propanoate ion = 0.0429 mol/L
Using the Henderson-Hasselbalch equation again:
pH = pKa + log([A-]/[HA])
pH = pKa + log([0.0429]/[0.00495])
pH = -log(1.3x10^-5) + log(0.0429/0.00495)
pH = -(-4.89) + log(8.672)
pH ≈ 4.89 + 0.94 (Approximately, because I'm not good at math!)
pH ≈ 5.83 (Or we can say, it's starting to neutralize, like my jokes!)
3. At 10 mL of KOH added: Following the same steps as before, we find that the pH will be approximately 6.69. (Or we can say, it's getting closer to neutral, just like those clowns who juggle knives!)
4. At equivalence point (when moles of propanoic acid = moles of KOH added): This is when the reaction is perfectly balanced, like a circus performer on a tightrope! At this point, all the propanoic acid has been neutralized, and we're left with only propanoate ions. So the pH would be determined by the concentration of propanoate ions, which is 0.300 mol/L since we've added 5 mL of 0.300 mol/L KOH.
5. At 1/2 of the equivalence point (when 1/2 of the moles of propanoic acid have reacted): This is like halfway through the show, a moment of suspense just like walking on broken glass! At this point, we've used up half of the propanoic acid, so we have half of the moles remaining. The concentration of propanoate ion would be half of 0.300 mol/L (which is 0.150 mol/L). Using the Henderson-Hasselbalch equation, we can calculate the pH.
6. At 20 mL of KOH added: Here, we would need to calculate the concentration of propanoate ion and use the Henderson-Hasselbalch equation to find the pH.
7. At 25 mL of KOH added: Again, we would need to calculate the concentration of propanoate ion and use the Henderson-Hasselbalch equation to find the pH.
Remember, pH is a measure of how acidic or basic a solution is. As the KOH is being added, it neutralizes the propanoic acid, gradually shifting the solution from acidic to basic. So be patient and enjoy the show as the pH levels dance their way up and down the scale!
To calculate the pH at various points during the titration of propanoic acid with KOH, we need to consider the reaction that occurs between the two substances, the balanced chemical equation, and the relevant chemical equilibrium expressions.
The balanced chemical equation for the reaction between propanoic acid (HA) and potassium hydroxide (KOH) is:
HA + KOH ⇌ K+ + A- + H2O
The relevant chemical equilibrium expression for the dissociation of propanoic acid is:
Ka = [A-][H3O+]/[HA]
At the beginning of the titration (0 mL added base), the solution contains only propanoic acid. Therefore, we need to calculate the initial concentration of propanoic acid (HA0) using the given volume (30.0 mL) and concentration (0.165 mol/L):
HA0 = (30.0 mL / 1000 mL) × 0.165 mol/L = 0.00495 mol
Using the balanced chemical equation, we can see that the initial concentration of A- is 0 mol/L, and the initial concentration of H3O+ is also 0 mol/L.
At the equivalence point (when stoichiometrically equal amounts of acid and base have reacted), the number of moles of propanoic acid (HA) will be equal to the number of moles of hydroxide ions (OH-). This allows us to determine the concentration of the hydroxide ions at that point:
OH- concentration = (0.300 mol/L) × (volume of base added at the equivalence point / 1000 mL)
Since propanoic acid is a weak acid, we need to consider that it undergoes partial dissociation. Thus, at the equivalence point, some of the propanoic acid will remain undissociated. Therefore, after the reaction, the concentration of A- will not be equal to the concentration of OH-. Instead, we need to calculate the concentrations of both species using the equilibrium expression for the dissociation of propanoic acid:
Ka = [A-][H3O+]/[HA]
At this point, we can use an ICE table (Initial, Change, Equilibrium) to determine the concentrations of the species:
Initial:
[HA] = HA0
[A-] = 0
[H3O+] = 0
Change:
[HA] = -x
[A-] = +x
[H3O+] = +x
Equilibrium:
[HA] = HA0 - x
[A-] = x
[H3O+] = x
Substituting these expressions into the equilibrium expression:
Ka = x * x / (HA0 - x)
Since the value of Ka is given as 1.3×10−5, we can rearrange the equation and solve for x to determine the concentration of H3O+ (which yields the pH):
x^2 = Ka * (HA0 - x)
x^2 + Ka * x - Ka * HA0 = 0
Solving this quadratic equation will give us the value of x, which can be used to determine the concentration of H3O+ at the equivalence point. Using the concentration of H3O+, we can calculate the pH.
To calculate the pH at 5, 10, 20, and 25 mL of added base, we can use the same method as described above. In these cases, the volume of base added will be different, and the calculations will involve different concentrations of acid and base.
It's important to note that these calculations assume the volume of the solution does not change during the titration and that the pKa of propanoic acid remains constant throughout the titration. Also, keep in mind that the pH will change abruptly near the equivalence point, which indicates a sharp change in the concentration of H3O+ ions.