Question

Find the ph of the equivalence points when 28.9 ml of 0.0850 m h2so3 is titrated with 0.0372 m naoh.

Answer 1

The pH at the equivalence point of the titration between 28.9 mL of 0.0850 M H2SO3 and 0.0372 M NaOH is approximately 8.25.

Step-by-step explanation:

The titration of a weak diprotic acid (H2SO3) with a strong base (NaOH) involves two equivalence points due to the two acidic protons in sulfuric acid. The first equivalence point corresponds to the complete neutralization of the first acidic proton, forming HSO3^- ions. The second equivalence point results from the neutralization of the second proton, forming SO3^2- ions. To find the pH at the second equivalence point:

Step 1: Determine the number of moles of H2SO3 present initially:

Moles = concentration × volume (in liters)

Moles of H2SO3 = 0.0850 mol/L × 0.0289 L = 0.00246 mol

Step 2: Calculate the moles of NaOH required to reach the second equivalence point:

The mole ratio between H2SO3 and NaOH is 1:2.

Moles of NaOH = 0.00246 mol H2SO3 × 2 mol NaOH / 1 mol H2SO3 = 0.00492 mol NaOH

Step 3: Determine the total volume at the second equivalence point:

Total volume = volume of H2SO3 + volume of NaOH

Total volume = 28.9 mL + V(NaOH) = 28.9 mL + (0.00492 mol / 0.0372 mol/L) = 28.9 mL + 0.1322 L = 0.1602 L

Step 4: Calculate the concentration of HSO3^- ions at the second equivalence point:

0.00246 mol of H2SO3 reacts to form 0.00246 mol of HSO3^-

Concentration of HSO3^- ions = moles / total volume = 0.00246 mol / 0.1602 L = 0.01534 M

Step 5: Calculate the pKa of the HSO3^- ion:

H2SO3 ⇌ H+ + HSO3^-

pKa = -log(Ka) = -log(1.7 × 10^-2) ≈ 1.77

Step 6: Calculate the pH at the second equivalence point:

pH = 1/2(pKa - log[HSO3^-]) = 1/2(1.77 - log[0.01534]) ≈ 8.25

Therefore, at the second equivalence point, the pH is approximately 8.25 due to the presence of the HSO3^- ions formed from the complete neutralization of the second acidic proton in H2SO3.

Answer 2

The pH of the equivalence point when titrating a weak diprotic acid, H2SO3, with a strong base, NaOH, is expected to be slightly above 7. This is due to the second dissociation of H2SO3, which has a pKa close to 7. Exact calculations would require more detailed chemical analysis.

Step-by-step explanation:

To find the pH of the equivalence point when 28.9 mL of 0.0850 M H2SO3 is titrated with 0.0372 M NaOH, we must consider the properties of the acid and base involved. H2SO3 (sulfurous acid) is a weak diprotic acid, which means it donates two protons per molecule in the reaction, and NaOH (sodium hydroxide) is a strong base. The equivalence point of a titration of a weak acid with a strong base is greater than 7 due to the formation of the conjugate base of the weak acid, which then hydrolyzes with water to form hydroxide ions.

In the case of sulfurous acid, the first dissociation (removal of the first proton) is more significant than the second. Thus, the pH at the equivalence point primarily depends on the second dissociation. The pKa values for H2SO3 are approximately 1.81 (first dissociation) and 7.21 (second dissociation). Since the second dissociation has a pKa close to 7, and the equivalence point pH is influenced by the weaker dissociation, we can anticipate the pH to be slightly above 7, unlike the equivalence point pH of a strong acid-strong base titration which is exactly 7.

To calculate the exact pH at the equivalence point, we would need to perform calculations involving the acid dissociation constant (Ka) of the second dissociation of H2SO3 and the concentration of the species in solution at the equivalence point. This would typically involve an ice table and Henderson-Hasselbalch equation, which is beyond the level of detail requested here. Without the specific calculations, we can reasonably estimate the pH to be slightly above 7 based on the properties of H2SO3.