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.
-600.0 mL of 1.00 M NaOH

Question

Quizplus · Accepted Answer

First, let's write the balanced equation for the reaction between the acidic H₂A and the basic NaOH: H₂A + NaOH → Na₂A + H2O Next, let's calculate how many moles of H₂A are present in the initial solution: moles of H₂A = M × L moles of H₂A = 1.00 M × 0.2000 L moles of H₂A = 0.2000 moles Now, we can use the balanced equation to calculate how many moles of NaOH are needed to titrate all of the acidic H₂A: 1 mole of H₂A reacts with 1 mole of NaOH moles of NaOH needed = moles of H₂A moles of NaOH needed = 0.2000 moles Finally, we can use the volume and concentration of NaOH added to calculate the new concentration of H₂A and then convert to pH using the equation for the dissociation of H₂A: H₂A + H2O ⇌ H+ + ₂A- pH = pKa + log([₂A-]/[H₂A]) First, let's calculate the moles of NaOH added: moles of NaOH added = M × L moles of NaOH added = 1.00 M × 0.6000 L moles of NaOH added = 0.6000 moles moles of H₂A remaining = moles of H₂A in initial solution - moles of NaOH added moles of H₂A remaining = 0.2000 moles - 0.6000 moles moles of H₂A remaining = -0.4000 moles (negative because we added more NaOH than needed) Since we added more NaOH than needed to titrate all of the acidic H₂A, we have a mixture of the acidic H₂A and its conjugate base ₂A-. To calculate the new concentration of H₂A and ₂A-, we need to use the Henderson-Hasselbalch equation: pH = pKa + log([₂A-]/[H₂A]) pKa = - log(Ka) = - log(1.0 × 10^-10) = 10 [H₂A] = moles of H₂A remaining / total volume [H₂A] = -0.4000 moles / 0.8000 L [H₂A] = 0.500 M [₂A-] = moles of NaOH added / total volume [₂A-] = 0.6000 moles / 0.8000 L [₂A-] = 0.750 M pH = 10 + log(0.750/0.500) pH = 13.40 Therefore, the correct answer is D, 13.40.

Calculate the PH When 200