During an acid-base titration, 25 mL of NaOH 0.2 M were required to neutralize 20 mL of HCl.

Calculate the pH of the solution for each:

a) Before titration
b) After adding 24.9 mL of NaOH
c) At equivalence pt
d) After adding 25. 1 mL of NaOH
e) The pH of the equivalence pt during a certain titration is 3.5
f) Which indicator would be adequate for this titration?
g) Which indicator would you suggest to be avoided? Explain

First, determine the M of the HCl. That will be 25*0.2/20 = 0.25M

Where is the equivalence point. That is at 25.0 mL, NaOH, of course.
The questions divide the titration curve up nicely into
a. at the beginning.
b. before the eq point.
c. at eq pt
d. just after eq pt

It's a matter of keeping account of the mols.
For a.
M HCl = 0.2M so pH = -log(HCl)
b.
millimols HCl initially = 5
millimols NaOH added = 4.98
millimols HCl remaining = 5-4.98 = 0.02
M HCl remaining = mmols/mL = 0.02/44.9
pH = -log(HCl)

c. At the eq you have NaCl and H2O. Neither Na^+ nor Cl^- are hydrolyzed; therefore, the pH is 7.0
d. mmols HCl initially = 5.0
mmols NaOH at 25.1 ix 25.1 x 0.2 = 5.02
mmols excess NaOH = 0.02
pH = mmols NaOH/mL solution = 0.02/45.1

I don't understand e.
Look in your text/notes for pH range of indicators.

a) Before titration, the pH of the solution is uncertain. It's like a mystery waiting to be solved, just like the plot of a detective novel. Who knows what secrets are hiding in that solution?

b) After adding 24.9 mL of NaOH, the pH of the solution will slightly increase. It'll be like a small burst of joy in an otherwise mundane day. Just a tiny step closer to reaching pH nirvana.

c) At the equivalence point, the pH of the solution will depend on the strength of the acid and base involved. It's like a balancing act between two skilled tightrope walkers, trying to find the perfect equilibrium. Will they succeed or fall into acidic or basic oblivion?

d) After adding 25.1 mL of NaOH, the pH of the solution will further increase. It's like getting a surprise extra scoop of ice cream on your dessert. Who can resist the sweet temptation of a slightly more basic solution?

e) If the pH of the equivalence point is 3.5, then it's like a punchline to a joke that only chemists would understand. It's a pH value that's neither too acidic nor too basic, like the perfect middle ground for comedy gold.

f) When choosing an indicator for this titration, something that changes color around the pH of 7 would be ideal. It's like having a chameleon that turns red when things get too acidic and blue when it gets too basic. Precisely, phenolphthalein would be a good choice.

g) I would suggest avoiding an indicator that changes color too far away from the equivalence point. For example, if we used litmus paper that turns blue at pH 10, it would be like trying to use a "Caution: Wet Floor" sign to warn people about an incoming tornado. It just wouldn't be accurate or helpful in this situation.

To calculate the pH at different points during the titration, we can use the equation:

pH = -log[H+]

First, let's determine the moles of acid and moles of base involved in the titration:

Moles of HCl = concentration of HCl × volume of HCl
Moles of HCl = 0.2 M × 20 mL / 1000 mL/L
Moles of HCl = 0.004 moles

Moles of NaOH = concentration of NaOH × volume of NaOH
Moles of NaOH = 0.2 M × 25 mL / 1000 mL/L
Moles of NaOH = 0.005 moles

a) Before titration:
The initial pH can be calculated by assuming the HCl solution does not contribute significantly to the pH. Therefore, the pH is primarily determined by the solvent, water. The pH of pure water at room temperature is usually around 7.

b) After adding 24.9 mL of NaOH:
To calculate the pH after adding 24.9 mL of NaOH, we need to determine the moles of excess NaOH.
Moles of NaOH remaining = moles of NaOH initially added - moles of HCl neutralized
Moles of NaOH remaining = 0.005 moles - 0.004 moles
Moles of NaOH remaining = 0.001 moles

Now, let's calculate the concentration of NaOH remaining:
Concentration of NaOH remaining = moles of NaOH remaining / volume of solution after addition
Concentration of NaOH remaining = 0.001 moles / (20 mL + 24.9 mL) / 1000 mL/L
Concentration of NaOH remaining = 0.001 moles / 44.9 mL / 1000 mL/L
Concentration of NaOH remaining = 0.022 M

Since NaOH is a strong base, it dissociates completely in water. Therefore, the OH- concentration will be equal to the remaining NaOH concentration.

pOH = -log[OH-] = -log[0.022]
pOH ≈ 1.66

pH = 14 - pOH = 14 - 1.66
pH ≈ 12.34

c) At the equivalence point:
The equivalence point is reached when the moles of acid and moles of base are in stoichiometric ratio, meaning they are completely neutralized. In this case, the moles of HCl and NaOH are equal.

Moles of NaOH required to neutralize HCl = Moles of HCl
0.005 moles = Moles of HCl

Since HCl is a strong acid, it dissociates completely in water. Therefore, the concentration of H+ will be equal to the concentration of HCl.

pH = -log[H+] = -log[0.2]
pH = 0.70

d) After adding 25.1 mL of NaOH:
To calculate the pH after adding 25.1 mL of NaOH, we need to determine the moles of excess NaOH.
Moles of NaOH remaining = moles of NaOH initially added - moles of HCl neutralized
Moles of NaOH remaining = 0.005 moles - 0.004 moles
Moles of NaOH remaining = 0.001 moles

Now, let's calculate the concentration of NaOH remaining:
Concentration of NaOH remaining = moles of NaOH remaining / volume of solution after addition
Concentration of NaOH remaining = 0.001 moles / (20 mL + 25.1 mL) / 1000 mL/L
Concentration of NaOH remaining = 0.001 moles / 45.1 mL / 1000 mL/L
Concentration of NaOH remaining = 0.022 M

Since NaOH is a strong base, it dissociates completely in water. Therefore, the OH- concentration will be equal to the remaining NaOH concentration.

pOH = -log[OH-] = -log[0.022]
pOH ≈ 1.66

pH = 14 - pOH = 14 - 1.66
pH ≈ 12.34

e) The pH of the equivalence point during a certain titration is 3.5:
At the equivalence point, the moles of acid and base are equal, indicating complete neutralization. For this particular titration, the pH at the equivalence point is 3.5.

f) Which indicator would be adequate for this titration:
To choose an appropriate indicator, we need to consider the pH range in which the stoichiometric point occurs. For this titration, the stoichiometric point occurs around pH 7. Therefore, an indicator with a pH range close to 7, such as phenolphthalein (pH range 8.3-10.0), would be suitable.

g) Which indicator would you suggest to be avoided? Explain:
Indicators with pH ranges far from the stoichiometric point should be avoided. For this titration, an indicator with a significant pH range below 7 or above 7 should be avoided. An example of an unsuitable indicator would be bromothymol blue (pH range 6.0-7.6) since its pH range does not adequately cover the stoichiometric point around pH 7.

To calculate the pH of the solution during the different stages of the titration, we need to understand the reaction between HCl and NaOH. The balanced chemical equation is as follows:

HCl + NaOH → NaCl + H2O

Given information:
- Volume of NaOH used: 25 mL
- Concentration of NaOH: 0.2 M
- Volume of HCl: 20 mL

a) Before titration:
Since there is no reaction between HCl and NaOH at this stage, the pH of the solution would be solely determined by the concentration of HCl. However, the concentration of HCl is not provided. Therefore, the pH value cannot be calculated without additional information.

b) After adding 24.9 mL of NaOH:
At this point, 24.9 mL of NaOH have been added, and the remaining 0.1 mL will be added in the next step. To calculate the pH, we need to determine the amount of excess HCl present in the solution.

The moles of HCl used can be calculated by multiplying the concentration (M) by the volume (L):
Moles of HCl = (20 mL/1000 mL) * (unknown M)

Since we have the moles of HCl used and the balanced equation, we can determine the moles of NaOH required for complete neutralization:
Moles of NaOH needed = moles of HCl used

Using the volume and concentration of the NaOH solution, we can calculate the moles of NaOH added:
Moles of NaOH added = (24.9 mL/1000 mL) * 0.2 M

To find the remaining moles of NaOH, subtract the added moles from the moles required for neutralization.

Next, calculate the concentration of excess HCl:
Excess HCl concentration = (moles of HCl / total volume of solution)

Finally, calculate the pH using the concentration of excess HCl.

c) At equivalence point:
The equivalence point occurs when the moles of acid (HCl) and base (NaOH) are stoichiometrically balanced. This means that equal moles of each reactant have been added.

Since equal moles of acid and base are present, the pH at the equivalence point can be determined by the dissociation of water:
H2O ↔ H+ + OH-

At the equivalence point, the concentrations of H+ and OH- are equal, resulting in pH = 7 (neutral pH).

d) After adding 25.1 mL of NaOH:
After adding 25.1 mL of NaOH, the moles of NaOH added can be calculated using the same approach as in previous steps.

Calculate the excess HCl concentration and then the pH using the concentration of excess HCl.

e) The pH of the equivalence point during a certain titration is 3.5:
Since the pH at the equivalence point is 3.5, this indicates that the solution is acidic. In order to reach an acidic pH at the equivalence point, the acid being titrated must be a weak acid, and the reaction between the acid and the base must be incomplete.

f) Which indicator would be adequate for this titration:
To choose an appropriate indicator for a titration, we should consider its pH range over which it changes color. For the given titration, which involves a strong acid (HCl) and a strong base (NaOH), a suitable indicator would be one with a pH range around the equivalence point (pH 7) and a color change within that pH range. Phenolphthalein would be suitable since it changes color in the pH range of 8.2 to 10.0, which is slightly above the equivalence point.

g) Which indicator would you suggest to be avoided, and why:
An indicator with a pH range below the equivalence point should be avoided in this titration. This is because it would not provide a clear color change around the equivalence point, making it difficult to determine the endpoint accurately. One example of an indicator to avoid for this titration would be methyl orange, which has a pH range of 3.1 to 4.4, well below the pH of the equivalence point (pH 7).