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A sample of 0.2140 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950 M NaOH. The acid required 30.0 mL of base to reach the equivalence point. After 15.0 mL of base had been added in the titration, the pH was found to be 6.50. What is the ��Ka for the unknown acid?

User Ashg
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Answer:

Step-by-step explanation:

To determine the Ka of the unknown acid, we can use the following steps:

Step 1: Calculate the number of moles of NaOH used.

moles of NaOH = Molarity × Volume of NaOH used

moles of NaOH = 0.0950 mol/L × 0.0300 L

moles of NaOH = 0.00285 mol

Step 2: Calculate the number of moles of acid used.

moles of acid = moles of NaOH (because they react in a 1:1 ratio)

moles of acid = 0.00285 mol

Step 3: Calculate the concentration of the acid.

the concentration of acid = moles of acid/volume of acid used

volume of acid used = volume of water + volume of NaOH used

volume of acid used = 25.0 mL + 30.0 mL = 55.0 mL = 0.0550 L

the concentration of acid = 0.00285 mol / 0.0550 L

concentration of acid = 0.0518 M

Step 4: Calculate the initial moles of acid in the solution.

moles of acid initially = concentration of acid × volume of water used

moles of acid initially = 0.0518 mol/L × 0.0250 L

moles of acid initially = 0.00130 mol

Step 5: Calculate the moles of acid that reacted from the initial amount to the endpoint.

moles of acid reacted = moles of acid initially - moles of acid at the endpoint

moles of acid reacted = 0.00130 mol - 0.00285 mol

moles of acid reacted = -0.00155 mol (negative because the acid was completely neutralized)

Step 6: Calculate the concentration of H+ ions at the equivalence point.

At the equivalence point, all the acid has been neutralized by the base, so the number of moles of H+ ions is equal to the number of moles of OH- ions. Therefore, the concentration of H+ ions is equal to that of OH- ions at the equivalence point.

moles of OH- ions = moles of NaOH used = 0.00285 mol

volume of solution at equivalence point = volume of water + volume of NaOH used

volume of solution at equivalence point = 25.0 mL + 30.0 mL = 55.0 mL = 0.0550 L

concentration of OH- ions = 0.00285 mol / 0.0550 L

the concentration of OH- ions = 0.0518 M

So, the concentration of H+ ions is also 0.0518 M at the equivalence point.

Step 7: Calculate the pKa of the acid.

The Henderson-Hasselbalch equation can be used to calculate the pKa of the acid:

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

At the equivalence point, the concentration of the acid is equal to the concentration of its conjugate base, so [A-] / [HA] = 1.

Therefore, pKa = pH + log(1) = pH

At the midpoint of the titration, the pH of the solution is equal to the pKa of the acid. According to the question, at the midpoint of the titration, the pH was 6.50. Therefore, the pKa of the acid is 6.50.

Step 8: Calculate the Ka of the acid.

The Ka of the acid can be calculated from the pKa using the formula:

Ka = 10^(-pKa)

Ka

User Matt Kuhns
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