Calculate the pH during the titration of 40 ml of 0.1000 M HCl with 0.1000 M NaOH solution after the following additions of base:

A.) 28.00 ML
B.) 39.80 ML
C.) 48.00 ML

40 mL of 0.1 M HCl obviously will take exactly 40 mL of 0.1 M NaOH to arrive at the equivalence point. So a and b are before the eq pt and c is after the eq pt.

For a.

HCl + NaOH ==> NaCl + H2O.
millimols HCl = 40 x 0.1 = 4
mmols NaOH = 28 x 0.1 = 2.8
Difference [which is HCl which is (H^+)] = 1.2 millimols.
M = mmols/mL (mols/L) = 1.2/(28+40) = ?
Then pH = -log(H^+)

b. Done the same way.

c. After the equivalence point you have a solution of NaCl in extra base.
millimols NaOH at 48 mL = 48 x 0.1 = 4.8
mmols HCl at the eq pt = 40 x 0.1 = 4.0
mmols NaOH in excess is 0.8
(OH^-) = 0.8/(48+40) = ?
pOH = -log(OH^-)
pH + pOH = pKw = 14. You know pKw and pOH, solve for pH.

Well, here we go with the titration circus! Let's calculate the pH during this wild act. Hang on tight!

A.) When you add 28.00 mL of 0.1000 M NaOH to 40 mL of 0.1000 M HCl, the number of moles of HCl initially was (0.040 L)(0.1000 mol/L) = 0.004 mol. The number of moles of NaOH added is (0.028 L)(0.1000 mol/L) = 0.0028 mol. Since we're adding a strong base to a strong acid, they will neutralize each other completely, forming water and a salt (NaCl).

The remaining HCl after the reaction would be (0.004 mol - 0.0028 mol) = 0.0012 mol. The total volume of the solution is (40 mL + 28 mL) = 68 mL = 0.068 L. So, the concentration of HCl in the solution is (0.0012 mol/0.068 L) = 0.0176 M.

Now, since HCl is a strong acid, it completely dissociates in water, meaning that the concentration of H+ ions is equal to the concentration of HCl: 0.0176 M.

To calculate the pH, we use the formula pH = -log[H+]. Therefore, pH = -log(0.0176) = 1.755.

So, during the first act of the titration circus, the pH is approximately 1.755. Let's move on to the next act!

B.) In the second act, when you add 39.80 mL of 0.1000 M NaOH, the number of moles of NaOH added is (0.0398 L)(0.1000 mol/L) = 0.00398 mol. The total volume of the solution is (40 mL + 39.80 mL) = 79.80 mL = 0.0798 L.

Since we have already established that the concentration of HCl is 0.0176 M, the remaining moles of HCl after the reaction would be (0.0012 mol - 0.00398 mol) = -0.00278 mol. Wait...there's something fishy going on here. A negative number of moles? That's not possible!

So, it seems like in this act, the NaOH has been added in excess and the reaction has gone beyond neutralization. Therefore, the pH will now be controlled by the excess of NaOH.

When NaOH reacts with water, it forms OH- ions. Since we have added an excess of NaOH, our solution will contain OH- ions. We can calculate the concentration of OH- by dividing the number of moles of NaOH added by the total volume of the solution: (0.00398 mol / 0.0798 L) = 0.0499 M.

To find the pOH, we use the formula pOH = -log[OH-]. Therefore, pOH = -log(0.0499) = 1.302.

Since pH + pOH = 14, we can calculate the pH using the equation pH = 14 - pOH: pH = 14 - 1.302 = 12.698.

So, in this second act, the pH is approximately 12.698. Let's move on to the final act!

C.) In the final act, when you add 48.00 mL of 0.1000 M NaOH, we can follow a similar procedure as in the previous act.

The number of moles of NaOH added is (0.048 L)(0.1000 mol/L) = 0.0048 mol. The total volume of the solution is (40 mL + 48 mL) = 88 mL = 0.088 L.

Again, since we have added an excess of NaOH, the reaction has gone beyond neutralization. The remaining moles of HCl after the reaction would be (0.0012 mol - 0.0048 mol) = -0.0036 mol. Another negative number of moles? That's wild!

So, just like in the previous act, the pH will now be controlled by the excess of NaOH. Using the same calculations, we find that the concentration of OH- is (0.0048 mol / 0.088 L) = 0.0545 M.

Calculating the pOH, we get pOH = -log(0.0545) = 1.262.

Using the equation pH = 14 - pOH, we find pH = 14 - 1.262 = 12.738.

Thus, in this third and final act, the pH is approximately 12.738.

And that's it, folks! The curtain falls on the pH calculations of this titration circus. I hope you enjoyed the show!

To calculate the pH during the titration, we need to determine the amount of acid and base remaining/reacted after each addition. We will use the following steps:

Step 1: Determine the initial amount of acid before any base is added.
Step 2: Calculate the moles of acid and moles of base for each addition.
Step 3: Determine the excess/remaining amount of acid and base after each addition.
Step 4: Calculate the resulting concentration of acid/base after each addition.
Step 5: Use the resulting concentration of acid/base to calculate the pH.

Let's begin with each case:

A) Addition of 28.00 mL of NaOH:

Step 1: Initial amount of acid:
Initial volume of HCl = 40 mL = 0.040 L
Initial moles of HCl = Volume x Molarity = 0.040 L x 0.1000 M = 0.004 moles

Step 2: Moles of acid and base:
Moles of NaOH = Volume x Molarity = 0.028 L x 0.1000 M = 0.0028 moles
Moles of HCl remaining = Initial moles - Moles of NaOH = 0.004 moles - 0.0028 moles = 0.0012 moles

Step 3: Remaining amount of acid and base:
Remaining volume of HCl = Initial volume - Volume of NaOH added = 40 mL - 28 mL = 12 mL = 0.012 L
Remaining moles of HCl = Moles of HCl remaining = 0.0012 moles
Remaining moles of NaOH = 0 moles (since all of it reacted with HCl)

Step 4: Concentration of acid and base:
Concentration of HCl = Remaining moles / Remaining volume = 0.0012 moles / 0.012 L = 0.1000 M
Concentration of NaOH = 0 moles / 0 L = 0 M (since all reacted)

Step 5: Calculate the pH:
The pH is determined by the concentration of H+ ions in the solution. Since the concentration of HCl is still 0.1000 M, the H+ concentration remains the same. Therefore, the pH remains unchanged.

B) Addition of 39.80 mL of NaOH:

Using the same steps as above, we get:
Remaining volume of HCl = 0.040 L - 0.0398 L = 0.0002 L
Concentration of HCl = 0.0012 moles / 0.0002 L = 6.00 M
Since the concentration of H+ ions changes, we can calculate the pH using the formula pH = -log[H+]:
pH = -log(6.00) = 0.221

C) Addition of 48.00 mL of NaOH:

Remaining volume of HCl = 0.040 L - 0.048 L = -0.008 L (Since we added more NaOH than the initial volume of HCl)
Therefore, there is no acid remaining, all the HCl has been neutralized by the added NaOH.
Since there are no acid molecules present, we cannot calculate the pH.

To calculate the pH during the titration, we need to determine the moles of acid and base at each addition and then use the balanced equation for the reaction between HCl and NaOH to find the moles of HCl and NaOH remaining.

Let's go step by step:

A.) 28.00 mL of NaOH added:
To find the moles of acid, we can use the formula:
moles = volume (L) x molarity (mol/L)

moles of HCl = 0.040 L x 0.1000 mol/L = 0.004 mol HCl
moles of NaOH = 0.028 L x 0.1000 mol/L = 0.0028 mol NaOH

From the balanced equation between HCl and NaOH:
HCl + NaOH → NaCl + H2O

We can see that the ratio between HCl and NaOH is 1:1, meaning that for every mole of HCl, we need one mole of NaOH to react completely.

Since we have 0.0028 mol NaOH, it means that 0.0028 mol HCl has reacted. Therefore, the moles of HCl remaining are 0.004 - 0.0028 = 0.0012 mol HCl.

Now, we can calculate the concentration of the remaining HCl:
concentration (mol/L) = moles / volume (L)

concentration of HCl = 0.0012 mol / (0.040 L + 0.028 L) = 0.017 mol/L

To calculate the pH, we can use the formula:
pH = -log[H+]

[H+] represents the concentration of H+ ions in the solution.

pH = -log(0.017) ≈ 1.77

Therefore, the pH after adding 28.00 mL of NaOH is approximately 1.77.

B.) 39.80 mL of NaOH added:
Following the same procedure:
moles of HCl = 0.004 mol
moles of NaOH = 0.0398 L x 0.1000 mol/L = 0.00398 mol
moles of HCl remaining = 0.004 - 0.00398 = 2.0 x 10^-5 mol
concentration of HCl = (2.0 x 10^-5 mol) / (0.040 L + 0.0398 L) = 2.5 x 10^-4 mol/L
pH = -log(2.5 x 10^-4) ≈ 3.60

Therefore, the pH after adding 39.80 mL of NaOH is approximately 3.60.

C.) 48.00 mL of NaOH added:
Following the same procedure:
moles of HCl = 0.004 mol
moles of NaOH = 0.048 L x 0.1000 mol/L = 0.0048 mol
moles of HCl remaining = 0.004 - 0.0048 = -0.0008 mol (negative because all HCl has reacted)
pH = -log(0) = Undefined (pH cannot be calculated when all acid has reacted)

Therefore, the pH after adding 48.00 mL of NaOH is undefined since all the acid has reacted.