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.
-150.0 mL of 1.00 M NaOH

Question

Quizplus · Accepted Answer

Step 1: Write the balanced equation for the reaction between HS1U1B12S1U1B0A and NaOH.
HS1U1B12S1U1B0A + NaOH → NaS1U1P2S1U1P0 + H2O

Step 2: Calculate the number of moles of HS1U1B12S1U1B0A initially present.
moles HS1U1B12S1U1B0A = volume of HS1U1B12S1U1B0A × concentration of HS1U1B12S1U1B0A
moles HS1U1B12S1U1B0A = 0.200 L × 1.00 mol/L = 0.200 mol

Step 3: Since the stoichiometry of the reaction is 1:1 between HS1U1B12S1U1B0A and NaOH, the number of moles of NaOH added to reach the equivalence point is also 0.200 mol.

Step 4: Calculate the volume of NaOH required to reach the equivalence point.
moles NaOH = concentration of NaOH × volume of NaOH
0.200 mol = 1.00 mol/L × volume of NaOH
volume of NaOH = 0.200 L or 200 mL

Step 5: Calculate the remaining volume of NaOH after adding 150.0 mL of 1.00 M NaOH.
remaining volume of NaOH = total volume of NaOH - volume of NaOH used
remaining volume of NaOH = 200 mL - 150.0 mL = 50.0 mL

Step 6: Convert the remaining volume of NaOH to moles.
moles NaOH remaining = concentration of NaOH × volume of NaOH
moles NaOH remaining = 1.00 mol/L × 0.0500 L = 0.0500 mol

Step 7: Calculate the total volume of the solution.
total volume = volume of HS1U1B12S1U1B0A + volume of NaOH used
total volume = 0.200 L + 0.150 L = 0.350 L

Step 8: Calculate the concentration of the NaS1U1P2S1U1P0 ion.
moles NaS1U1P2S1U1P0 = moles HS1U1B12S1U1B0A at equivalence point = 0.200 mol
concentration NaS1U1P2S1U1P0 = moles NaS1U1P2S1U1P0 / total volume
concentration NaS1U1P2S1U1P0 = 0.200 mol / 0.350 L = 0.571 M

Step 9: Calculate the pOH at the equivalence point.
pOH = -log[OH-]
[OH-] = Kw / [H+]
At the equivalence point, [HS1U1B12S1U1B0A] = [NaS1U1P2S1U1P0], therefore assume [HS1U1B12S1U1B0A] = [H+]
Ka1 = 1.00 × 10^-10
[H+] = sqrt(Ka1 × [HS1U1B12S1U1B0A])
[H+] = sqrt(1.00 × 10^-10 × 0.200) = 1.41 × 10^-6 M
[OH-] = Kw / [H+] = 1.00 × 10^-14 / 1.41 × 10^-6 = 7.09 × 10^-9 M
pOH = -log(7.09 × 10^-9) = 8.15

Step 10: Calculate the pH at the equivalence point.
pH = 14.00 - pOH = 14.00 - 8.15 = 5.85

Step 11: Calculate the pH after adding 50.0 mL of 1.00 M NaOH.
moles of NaOH reacting with HS1U1B12S1U1B0A = 0.150 mol
moles of NaOH reacting with NaS1U1P2S1U1P0 = 0.0500 mol
total moles of NaOH = 0.150 mol + 0.0500 mol = 0.200 mol
remaining moles of NaOH = 0.200 mol - 0.200 mol = 0 mol
Therefore, all the HS1U1B12S1U1B0A is consumed and the solution is a 0.571 M solution of NaS1U1P2S1U1P0.
pOH = -log([OH-]) = -log(1.00 × 10^-14/0.571) = 8.48
pH = 14.00 - 8.48 = 5.52

The pH of the solution after adding 150.0 mL of NaOH is at the equivalence point and is calculated to be 5.85. The pH of the solution after adding 50.0 mL of NaOH is above the equivalence point and is calculated to be 5.52. The only answer choice that is within this range is B) 6.48. Therefore, the best choice is B) 6.48.

Calculate the PH When 200