a 3.54 grams solid sample of an unknown monoprotic acid was dissolved in distilled water to produce a 47.0 mL solution at 25 degrees. This solution was then titrated with 0.2 M NaOH. The equivalence point was reached when 35.72 mL of 0.2 M NaOH was delivered.

A. find the number of moles of acid in the solid sample.

B. Find the molar mass of the unknown acid

C. Find Ka for the unknown acid if the pH at the equivalence point is 8.79

D. What was the pH after 63.2 mL of 0.2 M NaOH was delivered?

Tell us what you understand about this problem and what you don't understand. I can help you through that part. And please explain in detail what you don't understand.

10M

A. To find the number of moles of acid in the solid sample, we can use the formula:

moles of acid = (volume of NaOH used at equivalence point) x (molarity of NaOH)

moles of acid = 35.72 mL x 0.2 mol/L

Note: We need to convert the volume from milliliters to liters:

moles of acid = 0.03572 L x 0.2 mol/L

moles of acid = 0.007144 mol

Therefore, the number of moles of acid in the solid sample is 0.007144 mol.

B. To find the molar mass of the unknown acid, we can use the formula:

molar mass = (mass of acid sample) / (moles of acid)

molar mass = 3.54 g / 0.007144 mol

molar mass = 494.7 g/mol

Therefore, the molar mass of the unknown acid is 494.7 g/mol.

C. To find Ka for the unknown acid, we can use the expression:

pKa = pH at equivalence point - log(moles of acid / volume of solution at equivalence point)

We are given that the pH at the equivalence point is 8.79. Let's calculate pKa first:

pKa = 8.79 - log(0.007144 mol / (47.0 mL / 1000 mL/L))

Note: We convert the volume from milliliters to liters.

pKa = 8.79 - log(0.007144 mol / 0.0470 L)

pKa = 8.79 - log(0.1520)

Next, we can find Ka by taking the antilog (inverse log) of pKa:

Ka = 10^(-pKa)

Ka = 10^(-8.79)

Ka = 1.25 x 10^(-9)

Therefore, Ka for the unknown acid is 1.25 x 10^(-9).

D. To find the pH after 63.2 mL of 0.2 M NaOH was delivered, we can use the concept of the Henderson-Hasselbalch equation. The equation is:

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

At the equivalence point, all the acid has reacted with the base, so [A-] (conjugate base) is equal to moles of acid neutralized.

First, let's find the moles of acid neutralized:

moles of acid neutralized = (volume of NaOH used - volume at equivalence point) x (molarity of NaOH)

moles of acid neutralized = (63.2 mL - 35.72 mL) x 0.2 mol/L

moles of acid neutralized = 0.012456 mol

Now, we need to find the concentrations of [A-] and [HA]:

[A-] = moles of acid neutralized / (volume of solution - volume at equivalence point)

[A-] = 0.012456 mol / (47.0 mL / 1000 mL/L - 35.72 mL / 1000 mL/L)

[A-] = 0.012456 mol / (0.03328 L)

[A-] = 0.3740 M

[HA] = moles of acid / volume of solution

[HA] = 0.007144 mol / (47.0 mL / 1000 mL/L)

[HA] = 0.1520 M

Now, let's substitute these values into the Henderson-Hasselbalch equation:

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

pH = 8.79 + log(0.3740 M / 0.1520 M)

Finally, calculate the pH:

pH = 8.79 + log(2.4605)

pH = 8.79 + 0.3894

pH = 9.1794

Therefore, the pH after 63.2 mL of 0.2 M NaOH was delivered is approximately 9.18.

A. To find the number of moles of acid in the solid sample, we can use the formula:

moles of acid = volume of NaOH used * concentration of NaOH

In this case, the volume of NaOH used is 35.72 mL, and the concentration of NaOH is 0.2 M.

Convert the volume of NaOH used to liters:
35.72 mL = 35.72 mL * (1 L / 1000 mL) = 0.03572 L

Now we can calculate the moles of acid:
moles of acid = 0.03572 L * 0.2 M = 0.007144 moles

Therefore, there are 0.007144 moles of acid in the solid sample.

B. To find the molar mass of the unknown acid, we can use the formula:

molar mass = mass of acid / moles of acid

In this case, we know the mass of the acid is 3.54 grams, and we just calculated the moles of acid as 0.007144.

molar mass = 3.54 g / 0.007144 moles = 494.8 g/mol

Therefore, the molar mass of the unknown acid is 494.8 g/mol.

C. To find Ka for the unknown acid, we need to use the equation for the dissociation of the acid in water:

HA (acid) + H2O (water) ↔ H3O+ (hydronium ion) + A- (conjugate base)

At the equivalence point, the acid has completely reacted with the base, so we can assume that the concentration of the acid is equal to the concentration of the conjugate base. Let's call this concentration x.

The equation for the dissociation of the acid can be written as:

x^2 / (0.047 - x) = 10^-pH

where pH is 8.79.

Solving this equation will give us the value of x, which is the concentration of the acid and the conjugate base.

D. To find the pH after 63.2 mL of 0.2 M NaOH was delivered, we need to determine the concentration of the acid and the conjugate base. By using the same approach as in part C, we can calculate the concentration of acid and conjugate base at this point. Then, we can use the equation for pH:

pH = -log[H3O+]

Substituting the concentrations calculated in the previous step into this equation will give us the pH after 63.2 mL of 0.2 M NaOH was delivered.