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
A) 11.70
B) 10.00
C) 14.00
D) 13.40
E) none of these

Question

Quizplus · Accepted Answer

Step 1: Write the balanced chemical equation for the reaction between H₂A and NaOH. H₂A (aq) + NaOH (aq) → Na+ (aq) + _a1- (aq) + H2O (l) Step 2: Determine the initial moles of H₂A. moles of H₂A = M × V = 1.00 mol/L × 0.2000 L = 0.2000 mol Step 3: Determine the amount of NaOH added. moles of NaOH = M × V = 1.00 mol/L × 0.6000 L = 0.6000 mol Step 4: Determine the limiting reagent. NaOH is in excess, so H₂A is the limiting reagent. Step 5: Determine the moles of H₂A remaining after reaction with NaOH. moles of H₂A remaining = moles of H₂A initial - moles of NaOH used moles of H₂A remaining = 0.2000 mol - 0.6000 mol = -0.4000 mol (Note: A negative value indicates that all of the H₂A has reacted and there is excess NaOH remaining. In this case, we need to calculate the moles of excess NaOH and add it to the initial moles of NaOH to determine the total moles of NaOH used.) Step 6: Calculate the moles of excess NaOH. moles of excess NaOH = moles of NaOH added - moles of H₂A initial moles of excess NaOH = 0.6000 mol - 0.2000 mol = 0.4000 mol Step 7: Calculate the concentration of excess NaOH. excess NaOH concentration = moles of excess NaOH / volume of excess NaOH excess NaOH concentration = 0.4000 mol / 0.4000 L = 1.00 mol/L Step 8: Write the balanced chemical equation for the reaction between excess NaOH and water. NaOH (aq) + H2O (l) → Na+ (aq) + OH- (aq) Step 9: Determine the moles of excess NaOH that react with water. moles of excess NaOH used = moles of H2O formed = 0.4000 mol Step 10: Determine the concentration of OH- ions. OH- concentration = moles of excess NaOH used / volume of solution OH- concentration = 0.4000 mol / 0.8000 L = 0.500 mol/L Step 11: Determine the pOH. pOH = -log[OH-] = -log(0.500) = 0.3010 Step 12: Determine the pH. pH = 14.00 - pOH = 14.00 - 0.3010 = 13.70 Note: The pH is slightly lower than 14 because the H₂A molecule is a weak acid and the solution is a buffer. The addition of NaOH results in the formation of the conjugate base (_a1-) which reacts with water to form OH- ions, but the remaining H₂A can still act as a weak acid and partially neutralize the added base.

Calculate the PH When 200