Question

Consider the reaction of a 20.0 ml of 0.220 m c₅h₅nhcl (ka = 5.9 x 10⁻⁶) with 12.0 ml of 0.201 m csoh. with 0.00199 moles of c₅h₅nh⁺ and 0.00241 moles of c₅h₅n in the beaker, what would be the ph of this solution after the reaction goes to completion and the system reaches equilibrium?

Answer 1

To calculate the pH after the reaction of C5H5NHCl with CSOH, determine the excess OH- ions and use that to find the pOH, which is then used to calculate the pH. The pH will be basic as it's a weak acid-strong base titration.

Step-by-step explanation:

The question involves calculating the pH of a solution after a reaction between C5H5NHCl, a weak acid, and CSOH, a strong base, goes to completion. Given the reaction has taken place, we have 0.00199 moles of C5H5NH+ and 0.00241 moles of C5H5N in the beaker, with a Ka for C5H5NHCl being 5.9 x 10-6. After the neutralization reaction, excess OH- ions will determine the pH.

To find the pH, we need to calculate the concentration of OH- ions. From the given information, the number of moles of OH- can be found by subtracting the number of moles of C5H5NH+ from the number of moles of C5H5N, since the latter represents the amount of OH- that reacted. Then, we calculate the concentration of OH- ions by dividing the number of moles by the total volume of the solution after the reaction. Once the [OH-] is determined, we find the pOH using the formula pOH = -log[OH-]. Finally, we calculate the pH using the relationship pH = 14.00 - pOH.

It's important to note that the pH at the equivalence point of a weak acid-strong base titration is greater than 7, due to the production of the weak base during the neutralization reaction.

Answer 2

The pH of the solution after the reaction goes to completion and the system reaches equilibrium is 0.

Here's how to calculate the pH of the solution after the reaction goes to completion and the system reaches equilibrium:

1. Define initial concentrations and volumes:

C₅H₅NHCl concentration (c_c5h5nhcl) = 0.220 M

C₅H₅NHCl volume (v_c5h5nhcl) = 20.0 mL

CsOH concentration (c_csoh) = 0.201 M

CsOH volume (v_csoh) = 12.0 mL

2. Calculate initial moles of each species:

Moles of C₅H₅NHCl (n_c5h5nhcl) = c_c5h5nhcl * v_c5h5nhcl / 1000.0 = 0.0044 mol

Moles of CsOH (n_csoh) = c_csoh * v_csoh / 1000.0 = 0.002412 mol

3. Determine the limiting reactant:

Compare the ratios of initial moles of each reactant to their stoichiometric coefficients in the balanced equation:

C₅H₅NHCl + CsOH -> C₅H₅N + H₂O + Cl⁻

Ratio of C₅H₅NHCl to CsOH = 0.0044 mol / 0.002412 mol = 1.82

Stoichiometric ratio of C₅H₅NHCl to CsOH = 1:1

Since the ratio of C₅H₅NHCl to CsOH is greater than 1, C₅H₅NHCl is in excess and CsOH is the limiting reactan

4. Calculate the moles of each species after the reaction:

Moles of C₅H₅NH⁺ (n_c5h5nh_plus) = n_csoh = 0.002412 mol

Moles of C₅H₅N (n_c5h5n) = n_csoh = 0.002412 mol

Moles of OH⁻ (n_oh_minus) = 0 (all OH⁻ consumed by C₅H₅NH⁺)

5. Calculate the total volume of the solution:

Total volume (v_total) = v_c5h5nhcl + v_csoh = 20.0 mL + 12.0 mL = 32.0 mL

6. Calculate the concentrations of each species:

Concentration of C₅H₅NH⁺ (c_c5h5nh_plus) = n_c5h5nh_plus / v_total * 1000.0 = 0.002412 mol / 0.032 L * 1000.0 M/L = 0.0754 M

Concentration of C₅H₅N (c_c5h5n) = n_c5h5n / v_total * 1000.0 = 0.002412 mol / 0.032 L * 1000.0 M/L = 0.0754 M

Concentration of OH⁻ (c_oh_minus) = 0

7. Calculate the pKa of the conjugate acid:

pKa = -log10(Ka) = -log10(5.9 x 10⁻⁶) ≈ 5.23

8. Calculate the pOH:

pOH = -log10(c_oh_minus) = -log10(0) = ∞

9. Calculate the pH:

pH = 14.0 - pOH = 14.0 - ∞ = 0

Therefore, the pH of the solution after the reaction goes to completion and the system reaches equilibrium is 0.