Calculate the pH at the equivalence point for the titration of 0.25 M CH3COOH with 0.25
-5
M NaOH. (For CH3COOH, Ka= 1.8 â‹… 10 )
Well, pH at the equivalence point can be a real party... I mean, calculation! So, let me clown my way through this question.
First things first, we need to find the moles of CH3COOH and NaOH at the equivalence point. Since the concentration of both is 0.25 M and the volume is not given, let's just assume it's a total of 1 liter (just to make it simple).
Now, the balanced equation for the reaction is:
CH3COOH + NaOH → CH3COONa + H2O
According to the equation, 1 mole of CH3COOH reacts with 1 mole of NaOH to form 1 mole of CH3COONa and 1 mole of water.
So, at the equivalence point, both the moles of CH3COOH and NaOH will be the same. That means we have 0.25 moles of CH3COOH and 0.25 moles of NaOH.
To figure out the pH, we need to consider the dissociation of CH3COOH. The dissociation constant (Ka) value given is 1.8 x 10^-5.
Now, with all this information in our clowny toolbox, we can calculate the pH!
The dissociation of CH3COOH can be represented as:
CH3COOH ⇌ CH3COO- + H+
Using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
We know that at the equivalence point, [A-] = [HA], so we can simplify:
pH = pKa + log (1)
Since the log of 1 is 0, we get:
pH = pKa
Now, remembering that the value of Ka is 1.8 x 10^-5, we can calculate the pH:
pH = -log(1.8 x 10^-5)
And the grand reveal gives us the pH at the equivalence point for this titration of CH3COOH with NaOH...
*pause for dramatic effect*
pH = 4.74
So, there you have it! The pH at the equivalence point is 4.74, which is neither too acidic nor too basic. Just a party mode pH! Keep clowning around with chemistry, my friend!
To calculate the pH at the equivalence point of the titration of 0.25 M CH3COOH with 0.25 M NaOH, we need to determine the moles of acid and base present at the equivalence point.
1. Start by writing the balanced equation for the reaction:
CH3COOH + NaOH -> CH3COONa + H2O
2. Determine the number of moles of CH3COOH used in the titration:
moles CH3COOH = concentration x volume
moles CH3COOH = 0.25 M x volume (in liters)
3. Since the reactants are in a 1:1 ratio, the moles of NaOH added will be equal to the moles of CH3COOH used.
moles NaOH = moles CH3COOH
4. Use the moles of NaOH to calculate the concentration of NaOH in mol/L:
concentration NaOH = moles NaOH / volume (in liters)
concentration NaOH = moles CH3COOH / volume (in liters)
5. Using the equation for Ka of CH3COOH, calculate the concentration of H+ ions at the equivalence point:
Ka = [CH3COO-][H+]/[CH3COOH]
[H+] = Ka x [CH3COOH] / [CH3COO-]
At the equivalence point, [CH3COO-] = [CH3COOH], so we can substitute in the concentration of NaOH:
[H+] = Ka x (moles CH3COOH / volume (in liters)) / (moles CH3COOH / volume (in liters))
[H+] = Ka
6. Finally, calculate the pH:
pH = -log([H+])
Substitute the value of Ka to get the pH at the equivalence point.
To calculate the pH at the equivalence point for the titration of CH3COOH with NaOH, we need to determine the concentration of the CH3COOH and NaOH at the equivalence point.
At the equivalence point, the moles of acid and base are equal. This means that the moles of NaOH added is equal to the moles of CH3COOH initially present in the solution.
Given that the initial concentration of CH3COOH is 0.25 M and the Ka value is 1.8 x 10^-5, we can use the Henderson-Hasselbalch equation to find the pH. The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
pH is the pH at the equivalence point
pKa is the negative log of the acid dissociation constant Ka
[A-] is the concentration of the conjugate base (CH3COO-) at the equivalence point
[HA] is the concentration of the acid (CH3COOH) at the equivalence point
To find the concentration of CH3COO- at the equivalence point, we need to consider the reaction that occurs during the titration:
CH3COOH + NaOH -> CH3COO- + H2O
Since the moles of CH3COOH and NaOH are equal at the equivalence point, the concentration of CH3COOH will be halved. Therefore, at the equivalence point, the concentration of [CH3COOH] will be 0.25 M / 2 = 0.125 M.
Next, we need to calculate the concentration of CH3COO- at the equivalence point. Again, since the moles of CH3COOH and NaOH are equal, the concentration of CH3COO- will also be 0.125 M.
Now, we can substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = -log(1.8 x 10^-5) + log(0.125/0.125)
pH = -(-4.75) + log(1)
pH = 4.75 + 0
pH = 4.75
Therefore, the pH at the equivalence point for the titration of 0.25 M CH3COOH with 0.25 M NaOH is 4.75.
If we let HAc stand for acetic acid (CH3COOH), then the titration of acetic acid and NaOH is
HAc + NaOH ==>NaAc + H2O
At the equivalence point we have exactly neturalized HAc and NaOH and neither is in excess. All we have is the salt, NaAc, in water; therefore the pH is determined by the hydrolysis of the salt.
The (Ac^-) at the equivalence point is 0.125M (that's half the starting M of acetic acid and NaOH).
.......Ac^- + HOH ==> HAc + OH^-
I.....0.125............0.....0
C.......-x.............x.....x
E....0.125-x...........x.....x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.125-x)
Solve for x = (OH^-) and convert to pH.