You weigh a sample of a monoprotic unknown acid and dissolve it in 50.00 mL of distilled water. Exactly half of this solution is titrated with Sodium Hydroxide to the phenolphthalein end point.

The pH of the other half of the original solution is measured with a pH meter. The "neutralized" solution is added to the "original" solution and the pH of this combined "final" solution is also measured.
The following are the measured values:
Mass of unknown acid 1.2357 g
Volume of NaOH used in titration 16.04 mL
Concentration of the NaOH used 0.2098 M
pH of the original acid solution 2.02
pH of the final acid solution 3.29

CALCULATE the following
(a) Molecular Weight of Acid used in titration ___________________

(b) Molarity of UNKNOWN Acid solution from titration ___________________

(c) Ka of UNKNOWN Acid ___________________

(d) Concentration of undissociated Acid from pH measurements ___________________

(e) Total concentration of UNKNOWN acid from pH measurements ___________________

For each trial, enter the amount of heat lost by the calorimeter, qcalorimeter.

Be careful of the algebraic sign here and remember that the change in temperature is equal to the final temperature minus the initial temperature.
Report your answer using 4 digits. Note, this is 1 or 2 digits beyond the correct number of significant figures.

Trial # Initial Masswater Tiwater Tf qwater qcalorimeter
#1: 86.281 51.2 23.7 -9927.
#2: 87.777 49.0 29.0 -7345.
#3: 89.997 52.2 27.2 -9413.

To calculate the required values, we need to use various concepts from chemistry such as stoichiometry, acid-base titration, and the Henderson-Hasselbalch equation.

(a) Molecular Weight of Acid used in titration:
To find the molecular weight of the acid used in titration, we need to use the mass of the acid given and the molar ratio between the acid and sodium hydroxide. The balanced chemical equation for the reaction between the acid (HA) and NaOH is as follows:

HA + NaOH → NaA + H2O

From the equation, we can see that the molar ratio between the acid and NaOH is 1:1. Therefore, the number of moles of acid can be calculated using the following formula:

Moles of acid = Mass of acid / Molecular weight of acid

Solving for the molecular weight of the acid:

Molecular weight of acid = Mass of acid / Moles of acid

Given:
Mass of unknown acid = 1.2357 g

First, calculate the moles of acid:
Moles of acid = 1.2357 g / Molecular weight of acid

Since we do not yet know the molecular weight of the acid, we cannot accurately calculate the value at this point. Additional information is required.

(b) Molarity of UNKNOWN Acid solution from titration:
In an acid-base titration, the moles of acid reacting with the moles of the base can be equated by the following formula:

Moles of acid = Moles of base

We can calculate the moles of base (NaOH) using the volume and molarity provided:

Moles of NaOH = Volume of NaOH (L) × Concentration of NaOH (M)

Given:
Volume of NaOH used in titration = 16.04 mL = 0.01604 L
Concentration of NaOH used = 0.2098 M

Then, moles of NaOH = 0.01604 L × 0.2098 M

Since we have a 1:1 molar ratio between the acid and the base (according to the balanced equation from part (a)), the moles of acid can be calculated as:

Moles of acid = Moles of NaOH

Now, we can use the moles of acid and the volume of the solution to calculate the molarity of the unknown acid solution:

Molarity of unknown acid solution = Moles of acid / Volume of acid solution (L)

The volume of acid solution is half the volume of NaOH used, which is 0.00802 L (since half of 0.01604 L is used).

(c) Ka of UNKNOWN Acid:
The Ka value represents the acid dissociation constant. It can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log ([A-] / [HA])

Given:
pH of the original acid solution = 2.02
pH of the final acid solution = 3.29

By plugging in the pH values and rearranging the equation, we can solve for pKa:

pKa = pH - log ([A-] / [HA])

From the equation, we can see that the concentration of [A-] and [HA] are in the ratio of 1:1. Therefore, [A-] / [HA] = 1.

Substituting the given pH values:

pKa = 2.02 - log (1) for the original solution
pKa = 3.29 - log (1) for the final solution

(d) Concentration of undissociated Acid from pH measurements:
The pH of an acidic solution can be related to the concentration of the undissociated acid using the equation:

pH = -log ([HA])

Rearranging the equation gives us:

[HA] = 10^(-pH)

Given:
pH of the original acid solution = 2.02

Substituting the pH into the equation, we can calculate the concentration of the undissociated acid.

(e) Total concentration of UNKNOWN acid from pH measurements:
The total concentration of the unknown acid can be calculated using the equation:

[Total acid] = [HA] + [A-]

Since we know the concentration of the undissociated acid, [HA], from part (d), and the concentration of [A-] is equal to [HA], we can substitute these values into the equation to calculate the total concentration of the unknown acid.