a. A 0.1 molal solution of a weak monoprotic acid was found to depress the freezing point of water 0.1930C. Determine the Ka of the acid. You can assume 0.1 molal and 0.1 molar are equivalent.

b. 100 ml of a 0.1M solution of the above acid is titrated with 0.1 M NaOH. After an amount of NaOH is added the pH= 5.2. How much NaOH was added?

( c ) The titration is continued to equivalence. What is the pH at equivalence?
plzz help!

See my earlier post for the solution to a.

b is calculated from the Henderson-Hasselbalch e4quation.
c is calculated from the hydrolysis of the salt.
Show your work if you need additional help.

a. To determine the Ka of the weak monoprotic acid, we can use the freezing point depression method. The freezing point depression (∆Tf) is related to the molality (m) of the solution and the cryoscopic constant (Kf) of the solvent (water) by the equation:

∆Tf = Kf * m

Here, we are given ∆Tf = 0.1930°C and m = 0.1 molal. Since we assume 0.1 molal and 0.1 molar are equivalent, we can assume the molality is the same as the molarity.

Now, we need to find the cryoscopic constant (Kf) of water, which is 1.86°C/m.

Substituting the values into the equation:

0.1930°C = 1.86°C/m * 0.1 molal

Solving for m:

0.1930°C / (1.86°C/m) = 0.1 molal

Now, we can rearrange the equation for calculating molality:

m = moles of solute / kg of solvent

Since we know the solute is 0.1 moles of the weak monoprotic acid, we need to find the mass of the solvent (water) in kg.

To find the mass of 0.1 moles of water, we can multiply the molar mass of water (18 g/mol) by the number of moles:

mass of water = 0.1 moles * 18 g/mol = 1.8 g

Converting grams to kilograms:

mass of water = 1.8 g / 1000 = 0.0018 kg

Now, we can calculate the molality:

0.1 molal = 0.1 moles / 0.0018 kg

Simplifying the equation:

0.1 molal = 55.6 moles/kg

Finally, we can use the equation for Ka:

Ka = (Kw / [H+]) * [A-]

Since the weak monoprotic acid dissociates to [H+] and [A-], we need to find the concentration of [H+]. We can use the equation: [H+] = moles of acid / volume of solution.

However, we don't have the volume of the solution, so we need to first calculate it using the molality (55.6 moles/kg) and the mass of the solution.

Let's assume the mass of the solution is 1 kg for simplicity. Then the number of moles of the weak acid in the solution would be the same as the molality (0.1 molal). So, the concentration of [H+] can be calculated as:

[H+] = 0.1 mol / 1 L = 0.1 M

Since we assume 0.1 molal and 0.1 molar are equivalent, the concentration of [H+] is 0.1 M.

Therefore, the Ka of the weak acid can be calculated as:

Ka = (Kw / [H+]) * [A-] = (1.0 * 10^-14 / 0.1) * [A-]

We don't have information about the concentration of [A-], so we cannot determine the exact value of Ka without additional information.

b. To determine the amount of NaOH added, we can use the concept of pH. pH is calculated as the negative logarithm (base 10) of the concentration of H+ ions. In this case, the pH is given as 5.2.

pH = -log[H+]

To find the concentration of H+, we need to convert the pH value back into a concentration. We can use the formula:

[H+] = 10^(-pH)

So, [H+] = 10^(-5.2) = 5.01 * 10^(-6) M

Since NaOH is a strong base, it reacts with the weak monoprotic acid in a 1:1 ratio. So, the concentration of NaOH added is also 5.01 * 10^(-6) M.

To find the amount of NaOH added, we need to consider the volume of the solution. The volume is given as 100 mL, which is equal to 0.1 L.

Using the formula:

moles of NaOH = concentration of NaOH * volume of solution

moles of NaOH = (5.01 * 10^(-6) M) * (0.1 L)

Finally, we can calculate the amount of NaOH added in moles.

(c) To determine the pH at equivalence, we need to know the stoichiometry of the acid-base reaction. Assuming it is a 1:1 reaction, at equivalence, all of the acid will have reacted with the same amount of NaOH.

Thus, at equivalence, the number of moles of acid remaining will be zero, and the concentration of H+ will also be zero. Therefore, the pH at equivalence will be 7, which is neutral.