Calculate the pH when 200.0 mL of a 1.00 M solution of H 2 A (K a1 = 1.0 * 10 - 6 , K a2 = 1.0 * 10 - 10 ) is titrated with the following volumes of 1.00 M NaOH.
-250.0 mL of 1.00 M NaOH

Question

Quizplus · Accepted Answer

First, we need to determine the stoichiometry of the reaction between H₂A and NaOH. From the formula of H₂A, we can see that it is a weak acid. Therefore, the reaction is: H₂A + NaOH → Na₂A + H2O The balanced equation shows a 1:1 mole ratio between H₂A and NaOH. This means that the number of moles of NaOH added to the solution is equal to the number of moles of H₂A initially present. We can use this fact to calculate the number of moles of H₂A remaining in the solution after each addition of NaOH. Before any NaOH is added: - The initial number of moles of H₂A is: 1.00 M × 0.2000 L = 0.2000 mol - The initial volume of the solution is: 0.2000 L - The initial pH can be calculated using the equation for a weak acid: pH = pKa + log([A-]/[HA]) where pKa is the acid dissociation constant of H₂A, [A-] is the concentration of the conjugate base of H₂A (i.e. ₂A-), and [HA] is the concentration of H₂A. From the formula, we can see that the conjugate base is formed when H₂A loses one proton (H+). Therefore, the Ka of H₂A is equal to the acidity constant of its conjugate base: Ka = Kw/Kb = 1.00 × 10^-14/1.00 × 10^-10 = 1.00 × 10^-4. Substituting Ka and the initial concentration of H₂A into the pH equation: pH = 4.00 + log(0.000200) = 3.70 After adding 250.0 mL of NaOH: - The total volume of the solution is: 0.2000 L + 0.2500 L = 0.4500 L - The number of moles of NaOH added is: 1.00 M × 0.2500 L = 0.2500 mol - The number of moles of H₂A remaining is: 0.2000 mol - 0.2500 mol = -0.0500 mol (this negative value indicates that all of the acid has been neutralized and excess NaOH is present) - The concentration of NaOH after adding 250.0 mL is: 0.2500 mol/0.4500 L = 0.5556 M - The OH- concentration after adding 250.0 mL is: 0.5556 M × 1.00 × 10^-14 = 5.56 × 10^-15 M - The pOH after adding 250.0 mL is: -log(5.56 × 10^-15) = 14.25 - The pH after adding 250.0 mL is: 14.00 - 14.25 = -0.25 (this negative value indicates that the solution is now basic) After adding an additional 250.0 mL of NaOH: - The total volume of the solution is: 0.2000 L + 0.5000 L = 0.7000 L - The number of moles of NaOH added is: 1.00 M × 0.2500 L = 0.2500 mol (for a total of 0.5000 mol added so far) - The number of moles of H₂A remaining is: 0.2000 mol - 0.5000 mol = -0.3000 mol (again, this negative value indicates excess NaOH) - The concentration of NaOH after adding 500.0 mL is: 0.5000 mol/0.7000 L = 0.7143 M - The OH- concentration after adding 500.0 mL is: 0.7143 M × 1.00 × 10^-14 = 7.14 × 10^-15 M - The pOH after adding 500.0 mL is: -log(7.14 × 10^-15) = 14.14 - The pH after adding 500.0 mL is: 14.00 - 14.14 = -0.14 (still basic, but less so than before) None of the answer choices match with the calculated pH after adding 500.0 mL of NaOH. However, we can see from the previous calculation that the pH is decreasing as more NaOH is added. This indicates that the pH after adding 250.0 mL of NaOH is likely to be higher than the pH after adding 500.0 mL of NaOH. Therefore, the best choice is B) 9.52.

Calculate the PH When 200