Item 9
A 50.0 mL sample of 0.120 M HBr is titrated with 0.240 MNaOH.
Calculate the pH after the addition of the following volumes of
base.
Part A.)
0.0 mL
Express your answer using two decimal places.
Part B.)
20.0 mL
Express your answer using two decimal places.
Part C.)
24.9 mL
Express your answer using one decimal place.
Part D.)
25.0 mL
Express your answer using two decimal places.
Part E.)
25.1 mL
Express your answer using two decimal places.
Part F.)
40.0 mL
Express your answer using two decimal places.
Answer – We are given, [HBr] = 0.120 M , volume = 50.0 mL , [NaOH] = 0.240 M
Part A) pH after adding 0.0 mL of base –
When there is no any base is added then there is only acid and HBr is strong acid, so
[HBr] = [H+] = 0.120 M
We know ,
pH = -log [H+]
= - log 0.120
= 0.921
Part B) pH after adding 20.0 mL of base-
We know when NaOH is strong base and HBr also strong acid, so when then added to each other there is neutralization reaction as follow –
HBr + NaOH -----> NaBr + H2O
Calculation of moles –
Moles of HBr = 0.120 M * 0.050 L = 0.006 moles
Moles of NaOH = 0.240 M *0.020 L = 0.0048 moles
So , moles of acid is excess
Remaining moles of acid after reaction = 0.006 moles – 0.0048 moles
= 0.0012 moles
Total volume = 50+20 = 70 mL
So, new concentration of H+
[H+] = 0.0012 moles / 0.070 L
= 0.0171 M
So,
We know ,
pH = -log [H+]
= - log 0.0171
= 1.76
Part C) pH after adding 24.9 mL of base-
Calculation of moles –
Moles of HBr = 0.120 M * 0.050 L = 0.006 moles
Moles of NaOH = 0.240 M *0.0249 L = 0.00598 moles
So , moles of acid is excess
Remaining moles of acid after reaction = 0.006 moles – 0.00598 moles
= 2.4E-5 moles
Total volume = 50+24.9 = 74.9 mL
So, new concentration of H+
[H+] = 2.4E-5 moles / 0.0749 L
= 0.00032 M
So,
We know,
pH = -log [H+]
= - log 0.00032
= 3.49
Part D) pH after adding 25 mL of base-
Calculation of moles –
Moles of HBr = 0.120 M * 0.050 L = 0.006 moles
Moles of NaOH = 0.240 M *0.025 L = 0.006 moles
So, both acid and base reacted to each other completely and there is formed neutral pH
So pH = 7.0
Part E) pH after adding 25.1 mL of base-
Calculation of moles –
Moles of HBr = 0.120 M * 0.050 L = 0.006 moles
Moles of NaOH = 0.240 M *0.0251 L = 0.006024 moles
So , moles of base is excess
Remaining moles of base after reaction = 0.006024 moles – 0.0060 moles
= 2.4E-5 moles
Total volume = 50+25.1 = 75.1 mL
So, new concentration of OH-
[OH-] = 2.4E-5 moles / 0.0751 L
= 0.00032 M
So,
We know,
pOH = -log [OH-]
= - log 0.00032
= 3.49
pH = 14-pOH
= 14 – 3.49
= 10.5
Part F) pH after adding 40.0 mL of base-
Calculation of moles –
Moles of HBr = 0.120 M * 0.050 L = 0.006 moles
Moles of NaOH = 0.240 M *0.025 L = 0.006 moles
So, both acid and base reacted to each other completely and there is formed neutral pH
So pH = 7.0
Part E) pH after adding 25.1 mL of base-
Calculation of moles –
Moles of HBr = 0.120 M * 0.050 L = 0.006 moles
Moles of NaOH = 0.240 M *0.040 L = 0.0096 moles
So , moles of base is excess
Remaining moles of base after reaction = 0.0096 moles – 0.0060 moles
= 0.0036 moles
Total volume = 50+40 = 90 mL
So, new concentration of OH-
[OH-] = 0.0036 moles / 0.090 L
= 0.04 M
So,
We know,
pOH = -log [OH-]
= - log 0.04 M
= 1.40
pH = 14-pOH
= 14 – 1.40
= 12.6
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