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ID: 557063 • Letter: H
Question
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2. The following table is for a titration of 100.00 mL of 0.1000M HCl (analyte in flask) with 0.1000M NaOH (titrant in buret). Complete the table below. Based on your calculations in the table explain why there is a sudden change in pH surrounding the equivalence point of the titration. Show any necessary work. H1 (Molarity) H'] = 0.1000 M Volume of based added to the flask 0.00 mL 10.00 mL 90.00 mL 99.90 mL 99.99 mL 100.00 mL 100.01 mL 100.10 mL 101.00 mL 110.00 mL 1.00 1.09 Hint: Consider the dissociation of water Hint: Calculate excess [OH 8.7 10.7 11.7Explanation / Answer
HCl is a strong acid and NaOH is a strong base, means that a reaction between HCl and NaOH will be complete, once reaction takes place it will be NaCl and H2O. In the reaction almost no reverse equilibrium exists i.e. NaCl will not react with water and form HCl and NaOH.
In the titration, you are adding 0.1000 M NaOH to 100.00 mL of 0.1000 M HCl.
HCl + NaOH --> NaCl + H2O
………………………………………………………………………………………………………………………………………
No. of moles of HCl in 100 mL, 0.1000 M HCl = 0.1000 M x 100 mL/ 1000 mL = 0.01 mol.
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On adding 10.00 mL, 0.1000 M NaOH:
No. of moles of NaOH added = 0.1000 M x 10.00 mL/ 1000 mL = 0.001 mol
means 0.001 mol on NaOH will react with 0.001 mol of HCl from 0.01 mol.
So the amount of HCl left after the reaction = 0.01 mol - 0.001 mol = 0.009 mol.
During the addition, the volume of solution changed, 100.00 mL + 10.00 mL = 110.00 mL
After the addition of NaOH, the concentration of HCl will be 0.009 mol in 110.00 mL, the molarity will be 0.009 mol x 1000 mL / 110 mL = 0.08182 M
pH = -log[H+] = -log (0.08182) = 1.08715 = 1.09
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On adding 90.00 mL, 0.1000 M NaOH:
No. of moles of NaOH added = 0.1000 M x 90.00 mL/ 1000 mL = 0.009 mol
Amount of HCl left after the reaction = 0.01 mol - 0.009 mol = 0.001 mol.
Final volume = 100.00 mL + 90.00 mL = 190.00 mL
Molarity after addition = 0.001 mol x 1000 mL/190.00 mL = 0.005263 M
pH = -log[H+] = -log (0.005263) = 2.2787 = 2.28
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On adding 99.90 mL, 0.1000 M NaOH:
No. of moles of NaOH added = 0.1000 M x 99.90 mL/ 1000 mL = 0.00999 mol
Amount of HCl left after the reaction = 0.01 mol - 0.00999 mol = 0.00001 mol.
Final volume = 100.00 mL + 99.90 mL = 199.90 mL
Molarity after addition = 0.00001 mol x 1000 mL/199.90 mL = 5.0025 x 10^-5 M
pH = -log[H+] = -log (5.0025 x 10^-5) = 4.30
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On adding 99.99 mL, 0.1000 M NaOH:
No. of moles of NaOH added = 0.1000 M x 99.99 mL/ 1000 mL = 0.009999 mol
Amount of HCl left after the reaction = 0.01 mol - 0.009999 mol = 0.000001 mol.
Final volume = 100.00 mL + 99.99 mL = 199.99 mL
Molarity after addition = 0.000001 mol x 1000 mL/199.99 mL = 5.00025 x 10^-6 M
pH = -log[H+] = -log (5.00025 x 10^-6) = 5.30
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On adding 100.00 mL, all HCl reacts with NaOH, and form water and NaCl, here the self-ionization of water is the main source of H+ ions,
Ka of water = 10^-14, i.e. [H+][OH-] = 10^-14, [H+] = sqrt [10^-14] = 10^-7
[H+] = 10^-7 = pH = -log (10^-7) = 7
………………………………………………………………………………………………………………………………………
On adding 100.01 mL, 0.1000 M NaOH:
No. of moles of NaOH added = 0.1000 M x 100.01 mL/ 1000 mL = 0.010001 mol
Amount of HCl left after the reaction = 0.01 mol - 0.010001 mol = -0.000001 mol.
The negative value indicate that all HCl is used up and excess OH- ions are present,
[OH-] = 0.000001 mol
Final volume = 100.00 mL + 100.01 mL = 200.01 mL
Molarity of OH- after addition = 0.000001 mol x 1000 mL/200.01 mL = 4.99975 x 10^-6 M
Since [H+][OH-] = 10^-14,
[H+] = 10^-14/[OH-] = 10^-14/4.99975 x 10^-6 = 2.00000 x 10^-9 M
pH = -log[H+] = -log (2.00000 x 10^-9) = 8.6989 = 8.70
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On adding 100.10 mL, 0.1000 M NaOH:
No. of moles of NaOH added = 0.1000 M x 100.10 mL/ 1000 mL = 0.01001 mol
Amount of HCl left after the reaction = 0.01 mol - 0.01001 mol = -0.00001 mol.
[OH-] = 0.00001 mol
Final volume = 100.00 mL + 100.10 mL = 200.10 mL
Molarity of OH- after addition = 0.00001 mol x 1000 mL/200.10 mL = 4.99750 x 10^-5 M
[H+] = 10^-14/4.99750 x 10^-5 = 2.001 x 10^-10 M
pH = -log[H+] = -log (2.001 x 10^-10) = 9.6987 = 9.70
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On adding 101.00 mL, 0.1000 M NaOH:
No. of moles of NaOH added = 0.1000 M x 101.00 mL/ 1000 mL = 0.0101 mol
Amount of HCl left after the reaction = 0.01 mol - 0.0101 mol = -0.0001 mol.
[OH-] = 0.0001 mol
Final volume = 100.00 mL + 101.00 mL = 201.00 mL
Molarity of OH- after addition = 0.0001 mol x 1000 mL/201.00 mL = 4.97512 x 10^-4 M
[H+] = 10^-14/4.97512 x 10^-4 = 2.01 x 10^-11 M
pH = -log[H+] = -log (2.01 x 10^-11) = 10.6968 = 10.70
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On adding 110.00 mL, 0.1000 M NaOH:
No. of moles of NaOH added = 0.1000 M x 110.00 mL/ 1000 mL = 0.011 mol
Amount of HCl left after the reaction = 0.01 mol - 0.011 mol = -0.001 mol.
[OH-] = 0.001 mol
Final volume = 100.00 mL + 110.00 mL = 210.00 mL
Molarity of OH- after addition = 0.001 mol x 1000 mL/210.00 mL = 4.7619 x 10^-3 M
[H+] = 10^-14/4.7619 x 10^-3 = 2.10 x 10^-12 M
pH = -log[H+] = -log (2.10 x 10^-12) = 11.67778 = 11.7
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