Consider the titration of 40.0 mL of 0.400 M HI by 0.300 M NaOH.

A) Calculate the pH after 34.7 mL of NaOH has been added.
Answer:1.13

B)At what volume (in mL) of NaOH added does the pH of the resulting solution equal 7.00? Include the units of mL in your answer.

Oh, I see you're trying to titrate your way to a neutral solution. Well, let me put on my analytical clown nose and help you out!

To find the volume of NaOH needed to reach a pH of 7.00, we'll need to determine the number of moles of H+ ions remaining and then calculate the volume of NaOH required to neutralize it.

Given that you started with 40.0 mL of 0.400 M HI (hydroiodic acid), we can calculate the initial number of moles of H+ ions as follows:

Initial moles of H+ = initial concentration (M) × initial volume (L)
= 0.400 M × 0.0400 L
= 0.0160 moles

Now, we know that for every mole of NaOH added, one mole of H+ ions will be neutralized. In other words, the moles of H+ remaining will be equal to the moles of NaOH added. We'll use this information to calculate the pH at various volumes of NaOH until we reach pH 7.00.

So, here's the equation we'll use:

Remaining moles of H+ = initial moles of H+ - moles of NaOH added

At pH 7.00, we want the remaining moles of H+ to be zero:

0 = 0.0160 moles - moles of NaOH added

Now we solve for the moles of NaOH added:

moles of NaOH added = 0.0160 moles

Finally, we convert the moles of NaOH added to volume by using its concentration:

Volume (L) = moles of NaOH added / concentration (M)
= 0.0160 moles / 0.300 M
= 0.0533 L

Since 1 L = 1000 mL, we convert the volume to mL:

Volume (mL) = 0.0533 L × 1000 mL/L
= 53.3 mL

So, the volume of NaOH that needs to be added to reach pH 7.00 is 53.3 mL. Remember, though, your final answer should include the units of mL. Happy neutralizing!

To determine the volume (in mL) of NaOH added when the pH of the resulting solution is 7.00, we need to find the equivalence point of the titration. At the equivalence point, the moles of the acid (HI) will equal the moles of the base (NaOH).

First, let's calculate the moles of HI in the initial solution:
Moles of HI = Concentration of HI * Volume of HI
Moles of HI = (0.400 M) * (40.0 mL / 1000 mL/ L)
Moles of HI = 0.016 mol

Since HI reacts with NaOH in a 1:1 ratio, the moles of NaOH required to reach the equivalence point are also 0.016 mol.

Next, we need to determine the volume of NaOH required to react with 0.016 mol:
Volume of NaOH = Moles of NaOH / Concentration of NaOH
Volume of NaOH = 0.016 mol / 0.300 M
Volume of NaOH = 0.053 mL

Therefore, the volume of NaOH added when the pH of the resulting solution equals 7.00 would be 53 mL.

To calculate the pH after a certain volume of NaOH has been added in a titration, we need to understand the reaction that takes place and the stoichiometry involved.

The balanced chemical equation for the reaction between HI (hydroiodic acid) and NaOH (sodium hydroxide) is:

HI + NaOH → NaI + H2O

From the stoichiometry of the equation, we can see that the ratio of HI to NaOH is 1:1. This means that for every mole of NaOH added, one mole of HI is consumed.

Let's solve part A first:

A) Calculate the pH after 34.7 mL of NaOH has been added.

Step 1: Calculate the moles of NaOH added.
Given: Volume of NaOH = 34.7 mL
Concentration of NaOH = 0.300 M

Moles of NaOH = Volume (L) x Concentration (M)
= 34.7 mL x (1 L / 1000 mL) x 0.300 M
= 0.01041 moles

Step 2: Determine the moles of HI remaining.
Since the ratio of HI to NaOH is 1:1, the moles of HI remaining will be equal to the initial moles of HI minus the moles of NaOH added.

Initial moles of HI = Volume x Concentration
= 40.0 mL x (1 L / 1000 mL) x 0.400 M
= 0.016 moles

Moles of HI remaining = Initial moles of HI - Moles of NaOH added
= 0.016 moles - 0.01041 moles
= 0.00559 moles

Step 3: Calculate the concentration of HI.
Concentration of HI = Moles of HI remaining / Volume (L)
= 0.00559 moles / (40.0 mL x 1 L / 1000 mL)
= 0.1398 M

Step 4: Calculate the pH using the concentration of HI.
The pH of a solution can be calculated using the formula:
pH = -log[H+]

Since HI is a strong acid, it will completely ionize in water, and we can assume that the concentration of H+ ions is equal to the concentration of HI.

pH = -log[0.1398]
= 1.13

Therefore, the pH after 34.7 mL of NaOH has been added is 1.13.

Now, let's move on to part B:

B) At what volume (in mL) of NaOH added does the pH of the resulting solution equal 7.00?

To find the volume of NaOH at which the pH is 7.00, we need to determine the concentration of HI that would result in a pH of 7.00.

Step 1: Calculate the concentration of HI at pH 7.00.
Since pH is a measure of the concentration of H+ ions, we can convert the pH value to a concentration of H+ ions.

[H+] = 10^(-pH)
= 10^(-7.00)
= 1.0 x 10^(-7) M

This concentration represents the concentration of HI at the equivalence point.

Step 2: Determine the moles of NaOH needed to react with all the HI.
Since HI and NaOH react in a 1:1 ratio, the moles of NaOH needed will be equal to the initial moles of HI.

Moles of HI = Volume (L) x Concentration (M)
= 40.0 mL x (1 L / 1000 mL) x 0.400 M
= 0.016 moles

Moles of NaOH needed = 0.016 moles

Step 3: Calculate the volume of NaOH needed.
The volume (in mL) can be calculated using the formula:

Volume (mL) = Moles of NaOH needed / Concentration of NaOH (M) x 1000 ml / 1 L

Volume (mL) = 0.016 moles / 0.300 M x 1000 mL / 1 L
= 53.3 mL

Therefore, at a volume of 53.3 mL of NaOH added, the pH of the resulting solution would be 7.00.

millimoles HI = mL x M = 40.00 mL x 0.400 M = 16.00

millimoles NaOH added = 34.7 mL x 0.300 M = 10.41
..........................HI + NaOH ==> NaI + H2O
I.........................16.........0..............0.........0
add................................10.41.....................
C......................-10.41...-10.41.........+10.41
E........................ 5.59.........0..............10.41
So what do you have? That's an excess of 5.59 millimoles HI which is a strong acid and 10.41 mmols of NaI (the salt of a strong base and a strong acid which is neutal in solution). The pH will be determined by the HI concentration. Convert that to pH. Post your work if you get stuck.

B. For the pH to be 7.00 you will be at the equivalence point so you wantt the volume of the 0.300 M NaOH that will neutralize the 16.00 mmoles HI.
M NaOH = millimoles HI/mL NaOH. YOu know M = 0.300 and you know mmols HI = 16.00. Solve for pH = M = mmoles/mL