Consider the titration of 100.0 mL of 0.250 M aniline (K_b = 3.82 $\times$ 10^-¹⁰)with 0.500 M HCl.Calculate the pH of the solution at the stoichiometric point.

Question

Quizplus · Accepted Answer

Aniline is a weak base and HCl is a strong acid. The reaction between aniline and HCl is as follows: C6H5NH2 + HCl --> C6H5NH3+Cl- At the stoichiometric point, all the aniline has been reacted with the HCl. This means that all the aniline has been converted to C6H5NH3+ and the solution contains only the conjugate acid and chloride ions. Since C6H5NH3+ is the conjugate acid of a weak base, it is a weak acid and will not completely dissociate in water. Therefore, the pH at the stoichiometric point will be less than 7. To calculate the pH, we need to determine the amount of moles of aniline in the initial solution and the amount of moles of HCl added to reach the stoichiometric point. Then, we calculate the amount of moles of the conjugate acid produced at the stoichiometric point and use the dissociation constant of the conjugate acid to calculate the pH. Amount of moles of aniline = 0.250 mol/L x 0.1000 L = 0.0250 moles Amount of moles of HCl used at stoichiometric point = 0.500 mol/L x 0.1000 L = 0.0500 moles Amount of moles of C6H5NH3+ produced at stoichiometric point = 0.0250 moles Now, we can use the dissociation constant of C6H5NH3+ to calculate the pH: K_b = 3.82 F1F1F1S1F1F1F10 10^-¹⁰ K_b = [C6H5NH3+][OH-]/[C6H5NH2] K_b = [x][x]/[0.0250-x] K_b = x2/[0.0250-x] At equilibrium, the concentration of [OH-] = [C6H5NH3+] K_b = [OH-]2/[C6H5NH2] K_b = x2/0.0250 0.38 x 10^-10 = x2/0.0250 x = 1.7 x 10^-6 M pOH = -log[OH-] = -log(1.7 x 10^-6) = 5.77 pH = 14 - pOH = 8.23 Therefore, the pH of the solution at the stoichiometric point is 8.23, which corresponds to answer choice C.

Consider the Titration of 100 ×\times× 10-10)with 0

Consider the Titration of 100 $\times$ 10^-¹⁰)with 0