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
A) 10.00
B) 9.52
C) 8.00
D) 9.00
E) none of these

Question

Quizplus · Accepted Answer

The first step is to write out the balanced chemical equation for the reaction between H₂A and NaOH: H₂A + NaOH → NaH₂A + H2O In this reaction, one mole of NaOH reacts with one mole of H₂A to produce one mole of NaH₂A and one mole of water. Next, we need to determine the initial number of moles of H₂A present in the solution before any NaOH is added: moles H₂A = Molarity × Volume (in liters) moles H₂A = 1.00 M × 0.2000 L moles H₂A = 0.200 moles Now we can set up a table to track the reaction as NaOH is added: | | H₂A | NaOH | NaH₂A | |------|-----------------|--------|-----------------| | Before titration | 0.200 moles | 0 moles | 0 moles | | After adding 0.250 L NaOH | 0.200 moles - 0.250 moles | 0.250 moles | 0.250 moles | Note that at the equivalence point, the moles of H₂A will be equal to the moles of NaOH added. So we can set up an equation: moles H₂A = moles NaOH 0.200 - x = x 0.200 = 2x x = 0.100 Now we can calculate the concentration of NaH₂A at the equivalence point: Molarity NaH₂A = moles NaH₂A / volume NaOH added Molarity NaH₂A = 0.100 moles / 0.250 L Molarity NaH₂A = 0.400 M To calculate the pH, we need to determine the pKa of the conjugate acid, NaH₂A. We can do this by taking the negative logarithm of the acid dissociation constant, Ka: pKa = -log(Ka) pKa = -log(1.0 × 10^-10) pKa = 10 At the equivalence point, the solution contains equal amounts of NaH₂A and its conjugate base H₂A-. Since the pKa is 10, the pH will be slightly below 10 at the equivalence point. The best choice is B) 9.52, which is slightly below 10.

Calculate the PH When 200