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Acid-base titration calculation??
Can anyone help me on this?
I can never do them!
Can anyone run through this example with me?:
A 0.210mol dm-3 solution of potassium hydroxide was added from a burette to 25cm3 of a 0.160mol dm-3 solution of ethandioic acid in a conical flask.
Given that the value of the acid dissociation constant, Ka, for ethanoic acid is 1.74 x 10^-5, calculate the pH at 25 degrees of the solution in the conical flask at the following three points:
before any potassium hydroxide had been added (I can do this bit)
after 8cm3 of potassium hydroxide had been added
after 40cm3 of potassium hydroxide had been added
Thanks :]
I can never do them!
Can anyone run through this example with me?:
A 0.210mol dm-3 solution of potassium hydroxide was added from a burette to 25cm3 of a 0.160mol dm-3 solution of ethandioic acid in a conical flask.
Given that the value of the acid dissociation constant, Ka, for ethanoic acid is 1.74 x 10^-5, calculate the pH at 25 degrees of the solution in the conical flask at the following three points:
before any potassium hydroxide had been added (I can do this bit)
after 8cm3 of potassium hydroxide had been added
after 40cm3 of potassium hydroxide had been added
Thanks :]
Got the same answer as Eier above me. Although, more simply put:
1. Work out moles.
- You have volume and concentration of KOH, 8 and 40. So, work out moles using vol x concentration/1000, and you get 1.68x10^-3.
I worked out the [H+] was 1.67x10^-3, not sure if it's right. via [H+] = square root of Ka x [HA]
2. Deduct the OH- ions.
(1.68x10^-3) - (1.67x10^-3) = 1x10^-5.
pH = -log(1x10^-5) = 5. Sounds about right because it's a strong base VS a strong acid, so pH should increase.
======================================================
Just a quick question myself, the part at the bottom asking for 40cm3, is that asking for just 40 cm3, or is it asking for 40cm3 AND the previously added 8cm3?
Thanks!
1. Work out moles.
- You have volume and concentration of KOH, 8 and 40. So, work out moles using vol x concentration/1000, and you get 1.68x10^-3.
I worked out the [H+] was 1.67x10^-3, not sure if it's right. via [H+] = square root of Ka x [HA]
2. Deduct the OH- ions.
(1.68x10^-3) - (1.67x10^-3) = 1x10^-5.
pH = -log(1x10^-5) = 5. Sounds about right because it's a strong base VS a strong acid, so pH should increase.
======================================================
Just a quick question myself, the part at the bottom asking for 40cm3, is that asking for just 40 cm3, or is it asking for 40cm3 AND the previously added 8cm3?
Thanks!
Ok this is my method, I've never done a question like this so I pretty much made this up on the spot...
For the 1st q, you calculate the H+ ions, and acquire the pH. The initial H+ ion concn in the solution is 1.669x10^-3 moldm^-3.
For the 2nd q, when 8cm^3 of KOH (0.210moldm^-3) is added, find the moles of OH ions which = 8/1000 x 0.210 = 1.68x10^-3
Reacting ratio between H+ and OH- is 1:1 so the moles of H+ remaining after reaction is 1.669x10^-3 - 1.68x10^-3 is 0, there's an excess of OH- ions by 1.15x10^-3 mol.
The concentration of these OH ions is 1.15x10^-3 / (33/1000) <-- volume of soln after addition of further volume. Concn = 3.475x10^-4.
pOH = 3.46
pH + pOH = 14, so pH = 10.54
Same method for part 3 to get pH = 13.0
Note: I'm almost certainly wrong. xD When calculating the moles of KOH I've assumed the base is totally dissociated.
For the 1st q, you calculate the H+ ions, and acquire the pH. The initial H+ ion concn in the solution is 1.669x10^-3 moldm^-3.
For the 2nd q, when 8cm^3 of KOH (0.210moldm^-3) is added, find the moles of OH ions which = 8/1000 x 0.210 = 1.68x10^-3
Reacting ratio between H+ and OH- is 1:1 so the moles of H+ remaining after reaction is 1.669x10^-3 - 1.68x10^-3 is 0, there's an excess of OH- ions by 1.15x10^-3 mol.
The concentration of these OH ions is 1.15x10^-3 / (33/1000) <-- volume of soln after addition of further volume. Concn = 3.475x10^-4.
pOH = 3.46
pH + pOH = 14, so pH = 10.54
Same method for part 3 to get pH = 13.0
Note: I'm almost certainly wrong. xD When calculating the moles of KOH I've assumed the base is totally dissociated.
Irrelevance
Sounds about right because it's a strong base VS a strong acid.
It's a weak acid, ain't it? That's why you use Ka otherwise you'd just work it out normally.. besides, ethanedioic acid is diprotic, so it should have 2 Ka constants, 1 for the 1st dissociation, and another for the 2nd... but I guess the calculation doesnt really want you to think about it like that..
anyway, not so sure.. altho I think the method for my answer is right
. Just wait patiently for ChemistBoy to browse this thread.. lol.Creative
It's a weak acid, ain't it? That's why you use Ka otherwise you'd just work it out normally.. besides, ethanedioic acid is diprotic, so it should have 2 Ka constants, 1 for the 1st dissociation, and another for the 2nd... but I guess the calculation doesnt really want you to think about it like that..
anyway, not so sure.. altho I think the method for my answer is right. Just wait patiently for ChemistBoy to browse this thread.. lol.
anyway, not so sure.. altho I think the method for my answer is right
. Just wait patiently for ChemistBoy to browse this thread.. lol.Typo on my behalf, was thinking weak acid and typed strong. Looking good for the exam lol
The second part of this question is buffer law.
If you add 8 cm3 of a 0.210 M solution of KOH to 25cm3 of 0.16 M ( i.e. excess) ethanoic acid you will end up with a solutin containing both ethanoic acid and potassium ethanoate. This is a classic buffer mixture, weak acid plus the salt of a weak acid.
The reaction is
KOH + HA --> KA + H2O (1:1 mole equivalent)
Therefore mole of KOH reacted = moles of KA formed = 0.008 x 0.210 = 0.00168
Likewise this number of moles of HA have been used up
Therefore new moles of HA = initial - used up = 0.004 - 0.00168 = 0.00232
Now find the concentrations knowing the total volume is now 8 + 25 = 33cm3
concentration of HA = 0.00232/0.033 = 0.0703
concentration of KA (or A- if you prefer) = 0.05091
pH = pKa - log[HA]/[A-]
pH = 4.76 - 0.14 = 4.65
If you add 8 cm3 of a 0.210 M solution of KOH to 25cm3 of 0.16 M ( i.e. excess) ethanoic acid you will end up with a solutin containing both ethanoic acid and potassium ethanoate. This is a classic buffer mixture, weak acid plus the salt of a weak acid.
The reaction is
KOH + HA --> KA + H2O (1:1 mole equivalent)
Therefore mole of KOH reacted = moles of KA formed = 0.008 x 0.210 = 0.00168
Likewise this number of moles of HA have been used up
Therefore new moles of HA = initial - used up = 0.004 - 0.00168 = 0.00232
Now find the concentrations knowing the total volume is now 8 + 25 = 33cm3
concentration of HA = 0.00232/0.033 = 0.0703
concentration of KA (or A- if you prefer) = 0.05091
pH = pKa - log[HA]/[A-]
pH = 4.76 - 0.14 = 4.65
The last part is just calculation of moles of KOH in excess
Then concentration
then pH
excess moles KOH = total moles KOH added - total moles of acid = (0.040 x 0.21) - (0.025 x 0.16)
= 0.0084 - 0.004 = 0.0044 moles excess (this is now in 65cm3)
concentration of the KOH = 0.0044/0.065 = 0.0677 M
this is 0.0677 M in OH- ions
pOH = -log[OH-] = 1.169
pH = 14 - pOH = 12.83
Then concentration
then pH
excess moles KOH = total moles KOH added - total moles of acid = (0.040 x 0.21) - (0.025 x 0.16)
= 0.0084 - 0.004 = 0.0044 moles excess (this is now in 65cm3)
concentration of the KOH = 0.0044/0.065 = 0.0677 M
this is 0.0677 M in OH- ions
pOH = -log[OH-] = 1.169
pH = 14 - pOH = 12.83
charco
Now find the concentrations knowing the total volume is now 8 + 25 = 33cm3
concentration of HA = 0.00232/0.033 = 0.0703
concentration of KA (or A- if you prefer) = 0.05091
pH = pKa - log[HA]/[A-]
pH = 4.76 - 0.14 = 4.65
Now find the concentrations knowing the total volume is now 8 + 25 = 33cm3
concentration of HA = 0.00232/0.033 = 0.0703
concentration of KA (or A- if you prefer) = 0.05091
pH = pKa - log[HA]/[A-]
pH = 4.76 - 0.14 = 4.65
isn't the conc of HA 0.703 not 0.0703 so the pH is actually 3.619??
charco
The second part of this question is buffer law.
If you add 8 cm3 of a 0.210 M solution of KOH to 25cm3 of 0.16 M ( i.e. excess) ethanoic acid you will end up with a solutin containing both ethanoic acid and potassium ethanoate. This is a classic buffer mixture, weak acid plus the salt of a weak acid.
The reaction is
KOH + HA --> KA + H2O (1:1 mole equivalent)
Therefore mole of KOH reacted = moles of KA formed = 0.008 x 0.210 = 0.00168
Likewise this number of moles of HA have been used up
Therefore new moles of HA = initial - used up = 0.004 - 0.00168 = 0.00232
Now find the concentrations knowing the total volume is now 8 + 25 = 33cm3
concentration of HA = 0.00232/0.033 = 0.0703
concentration of KA (or A- if you prefer) = 0.05091
pH = pKa - log[HA]/[A-]
pH = 4.76 - 0.14 = 4.65
If you add 8 cm3 of a 0.210 M solution of KOH to 25cm3 of 0.16 M ( i.e. excess) ethanoic acid you will end up with a solutin containing both ethanoic acid and potassium ethanoate. This is a classic buffer mixture, weak acid plus the salt of a weak acid.
The reaction is
KOH + HA --> KA + H2O (1:1 mole equivalent)
Therefore mole of KOH reacted = moles of KA formed = 0.008 x 0.210 = 0.00168
Likewise this number of moles of HA have been used up
Therefore new moles of HA = initial - used up = 0.004 - 0.00168 = 0.00232
Now find the concentrations knowing the total volume is now 8 + 25 = 33cm3
concentration of HA = 0.00232/0.033 = 0.0703
concentration of KA (or A- if you prefer) = 0.05091
pH = pKa - log[HA]/[A-]
pH = 4.76 - 0.14 = 4.65
I'm pretty sure this is right, I forgot you made a buffer when you mixed these lol
.moreiniho
isn't the conc of HA 0.703 not 0.0703 so the pH is actually 3.619??
Initial moles of HA = 0.025 x 0.16 = 0.004 moles
Initial moles of KOH = 0.008 x 0.21 = 0.00168 moles
After reacting
Remaining moles of HA = 0.004 - 0.00168 = 0.00232
Moles of KA produced = 0.00168
new volume = 8 + 25 = 33 cm3
Conc HA = 0.00232/0.033 = 0.0703 M
Conc KA = 0.00168/0.033 = 0.0509 M
etc.
Original post by charco
The second part of this question is buffer law.
If you add 8 cm3 of a 0.210 M solution of KOH to 25cm3 of 0.16 M ( i.e. excess) ethanoic acid you will end up with a solutin containing both ethanoic acid and potassium ethanoate. This is a classic buffer mixture, weak acid plus the salt of a weak acid.
The reaction is
KOH + HA --> KA + H2O (1:1 mole equivalent)
Therefore mole of KOH reacted = moles of KA formed = 0.008 x 0.210 = 0.00168
Likewise this number of moles of HA have been used up
Therefore new moles of HA = initial - used up = 0.004 - 0.00168 = 0.00232
Now find the concentrations knowing the total volume is now 8 + 25 = 33cm3
concentration of HA = 0.00232/0.033 = 0.0703
concentration of KA (or A- if you prefer) = 0.05091
pH = pKa - log[HA]/[A-]
pH = 4.76 - 0.14 = 4.65
If you add 8 cm3 of a 0.210 M solution of KOH to 25cm3 of 0.16 M ( i.e. excess) ethanoic acid you will end up with a solutin containing both ethanoic acid and potassium ethanoate. This is a classic buffer mixture, weak acid plus the salt of a weak acid.
The reaction is
KOH + HA --> KA + H2O (1:1 mole equivalent)
Therefore mole of KOH reacted = moles of KA formed = 0.008 x 0.210 = 0.00168
Likewise this number of moles of HA have been used up
Therefore new moles of HA = initial - used up = 0.004 - 0.00168 = 0.00232
Now find the concentrations knowing the total volume is now 8 + 25 = 33cm3
concentration of HA = 0.00232/0.033 = 0.0703
concentration of KA (or A- if you prefer) = 0.05091
pH = pKa - log[HA]/[A-]
pH = 4.76 - 0.14 = 4.65
Hi, this may be really late to ask but why did you do this step 'concentration of KA (or A- if you prefer) = 0.05091' ?
Thanks
Original post by coconut64
Hi, this may be really late to ask but why did you do this step 'concentration of KA (or A- if you prefer) = 0.05091' ?
Thanks
Thanks
I wrote this over 8 years ago!!
But the answer remains the same.
Read the post!
Original post by charco
I wrote this over 8 years ago!!
But the answer remains the same.
Read the post!
But the answer remains the same.
Read the post!
Thanks, I just realised I haven't learnt buffers yet and that is applied in the question,
Original post by charco
The last part is just calculation of moles of KOH in excess
Then concentration
then pH
excess moles KOH = total moles KOH added - total moles of acid = (0.040 x 0.21) - (0.025 x 0.16)
= 0.0084 - 0.004 = 0.0044 moles excess (this is now in 65cm3)
concentration of the KOH = 0.0044/0.065 = 0.0677 M
this is 0.0677 M in OH- ions
pOH = -log[OH-] = 1.169
pH = 14 - pOH = 12.83
Then concentration
then pH
excess moles KOH = total moles KOH added - total moles of acid = (0.040 x 0.21) - (0.025 x 0.16)
= 0.0084 - 0.004 = 0.0044 moles excess (this is now in 65cm3)
concentration of the KOH = 0.0044/0.065 = 0.0677 M
this is 0.0677 M in OH- ions
pOH = -log[OH-] = 1.169
pH = 14 - pOH = 12.83
I know it's been a VERY long time, but I needed this now and your answer was well structured so I understood this question easily after many attempts on my own. So thank you so much!
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