# Buffer solutions

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#1
So in my textbook it says "when mixing equal quantities each original concentration will be halved" (of acid and salt) but this is within an example so I'm not sure if this always applies? How would I know the proportion to increase/decrease the concentration of the acid or salt solution by within my calculation if 'equal amounts of each' are added?

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7 years ago
#2
It is just a dilution. Assuming you mixed Vacid and Vsalt new total volume is Vacid+Vsalt and new concentration (of a substance) is

If Vacid=Vsalt fraction will equal 1/2. For other volumes it will have some other value.
0
7 years ago
#3
(Original post by Branny101)
So in my textbook it says "when mixing equal quantities each original concentration will be halved" (of acid and salt) but this is within an example so I'm not sure if this always applies? How would I know the proportion to increase/decrease the concentration of the acid or salt solution by within my calculation if 'equal amounts of each' are added?

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Say solution 1 has a concentration of 50 moles per ml. Solution 2 has a concentration of 25 moles per ml.

Adding each solution in equal amount would give:

A 1 ml solution that has a concentration of 25 moles per ml for solution 1 and 12.5 moles per ml for solution 2

- Concentration is 'amount per unit volume' which is usually translated as milli moles per milli litre (ml). To simplify calculations, assume your solution is out of 1 ml or 1 litre, as concentration units are either moles per litre or milli moles per ml

Assuming constant concentration:
What is the amount (in moles) in half this ml? half the amount in 1 ml
What is the amount (in moles) in 2 ml? double the amount in 1 ml
What is the amount (in moles) in 10 ml? ten times the amount in 1 ml

Concentrations of solution 1 (same stock itself, nothing else added) in:
half a ml of solution 1? same as in 1 ml
2 ml of solution 1? same as in 1 ml
10 ml of solution 1? same as in 1 ml

That is because your molecule (e.g. NaCl) is floating around stochastically (a big word meaning randomly while following a certain pattern) in your little tube half-filled with your solution. It has the same concentration all over the solution, as if you add sugar to your tea wherever you drink you feel the same intensity of the sugar. In just 10% of the tea in your mug, you have 10% of the amount of sugar you originally added to your tea while this 10% has the same concentration as in the rest of the mug.

Now assuming we add water (or buffer or solution 2):

What is the concentration in half this ml while the other half is water? Half the concentration in 1 ml
What is the concentration in 250 ul (micro litre) solution 1 and 750 ul water? 1 quarter of the concentration in 1 ml

What is the concentration in 10 ul (solution 1) and 990 ul water?

For this you would have to do a small calculation. The concentration ratio is the amount of solution 1 divided by the total volume i.e. 10 divided by 1000 (990 + 10) which would give you a 0.01 or 1 over a hundred. Meaning a 1 to 100 ratio or for every mole of solution 1, there are 100 moles of water.

So in your example you made a solution which is 50% acid and 50% salt. To simply calculations we can assume it is out of a 1 ml, you added 500 ul solution 1 and 500 ul solution 2. Doing the calculation of 500 divided by 1000 would give you 0.5 or 1 / 2 or one in two ratio.

- mole per litre is M. milli mole per ml is M. milli mole per litre is mM. This is the usual concentration units we use in biochemistry, not moles per litre. Though whenever you get confused try knowing that concentration is moles per 1 litre.

Hope this cleared some cloud
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#4
(Original post by Borek)
It is just a dilution. Assuming you mixed Vacid and Vsalt new total volume is Vacid+Vsalt and new concentration (of a substance) is

If Vacid=Vsalt fraction will equal 1/2. For other volumes it will have some other value.
I'm confused ahaha, sorry

(Original post by Masoudy)
Say solution 1 has a concentration of 50 moles per ml. Solution 2 has a concentration of 25 moles per ml.

Adding each solution in equal amount would give:

A 1 ml solution that has a concentration of 25 moles per ml for solution 1 and 12.5 moles per ml for solution 2

- Concentration is 'amount per unit volume' which is usually translated as milli moles per milli litre (ml). To simplify calculations, assume your solution is out of 1 ml or 1 litre, as concentration units are either moles per litre or milli moles per ml

Assuming constant concentration:
What is the amount (in moles) in half this ml? half the amount in 1 ml
What is the amount (in moles) in 2 ml? double the amount in 1 ml
What is the amount (in moles) in 10 ml? ten times the amount in 1 ml

Concentrations of solution 1 (same stock itself, nothing else added) in:
half a ml of solution 1? same as in 1 ml
2 ml of solution 1? same as in 1 ml
10 ml of solution 1? same as in 1 ml

That is because your molecule (e.g. NaCl) is floating around stochastically (a big word meaning randomly while following a certain pattern) in your little tube half-filled with your solution. It has the same concentration all over the solution, as if you add sugar to your tea wherever you drink you feel the same intensity of the sugar. In just 10% of the tea in your mug, you have 10% of the amount of sugar you originally added to your tea while this 10% has the same concentration as in the rest of the mug.

Now assuming we add water (or buffer or solution 2):

What is the concentration in half this ml while the other half is water? Half the concentration in 1 ml
What is the concentration in 250 ul (micro litre) solution 1 and 750 ul water? 1 quarter of the concentration in 1 ml

What is the concentration in 10 ul (solution 1) and 990 ul water?

For this you would have to do a small calculation. The concentration ratio is the amount of solution 1 divided by the total volume i.e. 10 divided by 1000 (990 + 10) which would give you a 0.01 or 1 over a hundred. Meaning a 1 to 100 ratio or for every mole of solution 1, there are 100 moles of water.

So in your example you made a solution which is 50% acid and 50% salt. To simply calculations we can assume it is out of a 1 ml, you added 500 ul solution 1 and 500 ul solution 2. Doing the calculation of 500 divided by 1000 would give you 0.5 or 1 / 2 or one in two ratio.

- mole per litre is M. milli mole per ml is M. milli mole per litre is mM. This is the usual concentration units we use in biochemistry, not moles per litre. Though whenever you get confused try knowing that concentration is moles per 1 litre.

Hope this cleared some cloud
Ooooooh okay buy say if you are given a concentration so you couldn't assume am amount. Let's say for example 0.5moldm3 of benzoic acid and 0.25 moldm3 of sodium benzoate for a buffer solution and I were told that they were added in equal amounts ( which you say is 50/50) how would the numbers add up?!

Thank you both kindly for your help btw!

EDIT: Say that we were asked with the above values to work out the pH of the buffer solution given that benzoic acid has a Ka value of 6.4x10-5. I understand the stages after finding the solutions because that's just plugging in, its just that bit you were explaining...

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7 years ago
#5
(Original post by Masoudy)
Say solution 1 has a concentration of 50 moles per ml. Solution 2 has a concentration of 25 moles per ml.
Good luck finding such solutions. Even 50 moles per liter is extremely high.

That is because your molecule (e.g. NaCl) is floating around stochastically (a big word
and completely unnecessary one.

meaning randomly while following a certain pattern
That's not what it means. Either randomly, or following a pattern.

It has the same concentration all over the solution
That's all that is important here.

Branny101: have you visited the link from my post? Linked page says basically the same thing Masoudy tried to say - dilutions are calculated from the mass conservation. Whatever you put into the solution stays there. Calculate how much substance you have put into the mixture, and it won't change. The only thing that changes is the volume of the solution. Assume the final volume is sum of the volumes of solutions mixed (that's not always true, but is usually close to true) and use this new volume and amount of substance to calculate new concentration.
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#6
(Original post by Borek)
Good luck finding such solutions. Even 50 moles per liter is extremely high.

and completely unnecessary one.

That's not what it means. Either randomly, or following a pattern.

That's all that is important here.

Branny101: have you visited the link from my post? Linked page says basically the same thing Masoudy tried to say - dilutions are calculated from the mass conservation. Whatever you put into the solution stays there. Calculate how much substance you have put into the mixture, and it won't change. The only thing that changes is the volume of the solution. Assume the final volume is sum of the volumes of solutions mixed (that's not always true, but is usually close to true) and use this new volume and amount of substance to calculate new concentration.
Like I understand the theory but struggle to apply it. The questions I'm given are usually given in the form of my edit above. Or they'll say something like 0.1moldm-3 of NaOH was added to the solution and half reacted, what is now the pH of the solution. But considering the question above Borek - how would I answer it?

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7 years ago
#7
(Original post by Branny101)
I'm confused ahaha, sorry

Ooooooh okay buy say if you are given a concentration so you couldn't assume am amount. Let's say for example 0.5moldm3 of benzoic acid and 0.25 moldm3 of sodium benzoate for a buffer solution and I were told that they were added in equal amounts ( which you say is 50/50) how would the numbers add up?!

Thank you both kindly for your help btw!

EDIT: Say that we were asked with the above values to work out the pH of the buffer solution given that benzoic acid has a Ka value of 6.4x10-5. I understand the stages after finding the solutions because that's just plugging in, its just that bit you were explaining...

Posted from TSR Mobile
0.25 mol dm3 benzoic acid
0.125 mol dm3 sodium benzoate

in the new buffer solution. You see what I did? Just divided both concentrations by half as you are mixing half of the first solution with half of the second one, giving half of both concentrations in a single solution.

It is like diluting both solutions by 50%; they dilute each other

Relevant example: you had 0.5 mol dm3 benzoic acid and you added water in equal amount. You would get 0.25 mol dm3 benzoic acid after adding the water as it would be 50% less concentrated.
0
7 years ago
#8
(Original post by Borek)
Good luck finding such solutions. Even 50 moles per liter is extremely high.

and completely unnecessary one.

That's not what it means. Either randomly, or following a pattern.

That's all that is important here.

Branny101: have you visited the link from my post? Linked page says basically the same thing Masoudy tried to say - dilutions are calculated from the mass conservation. Whatever you put into the solution stays there. Calculate how much substance you have put into the mixture, and it won't change. The only thing that changes is the volume of the solution. Assume the final volume is sum of the volumes of solutions mixed (that's not always true, but is usually close to true) and use this new volume and amount of substance to calculate new concentration.
Borek I do this chemical mixing in real life. Not sure why you are criticising
0
#9
(Original post by Masoudy)
0.25 mol dm3 benzoic acid
0.125 mol dm3 sodium benzoate

in the new buffer solution. You see what I did? Just divided both concentrations by half as you are mixing half of the first solution with half of the second one, giving half of both concentrations in a single solution.

It is like diluting both solutions by 50%; they dilute each other

Relevant example: you had 0.5 mol dm3 benzoic acid and you added water in equal amount. You would get 0.25 mol dm3 benzoic acid after adding the water as it would be 50% less concentrated.
Thank you. If I'm not mistakening this then it should be really simple

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7 years ago
#10
(Original post by Borek)
It is just a dilution. Assuming you mixed Vacid and Vsalt new total volume is Vacid+Vsalt and new concentration (of a substance) is

If Vacid=Vsalt fraction will equal 1/2. For other volumes it will have some other value.
This, just understand that concentration is a measurement of the number of moles in a set volume (hence it has units of moles ml-1). If you increase the volume by adding another substance then clearly the concentration will also decrease.

The equation Broek has given is an easy way of quantifying this, it is something that you need to learn.
0
7 years ago
#11
(Original post by Masoudy)
Borek I do this chemical mixing in real life.
You mean you ever prepared 50 mole per mL solution, or am I misunderstanding you?

If so, can you share the recipe?
0
7 years ago
#12
(Original post by Borek)
You mean you ever prepared 50 mole per mL solution, or am I misunderstanding you?

If so, can you share the recipe?
The above is a simple example that is understood easier than mmoles per ml

Its the principle that he is asking. Thats why he didn't understand what you wrote even though your equation is more accurate than mine.
0
7 years ago
#13
(Original post by Masoudy)
The above is a simple example that is understood easier than mmoles per ml
Never use examples that don't follow reality. They do more harm than good. 1 mol/1 L would do the trick.
0
#14
(Original post by Borek)
Never use examples that don't follow reality. They do more harm than good. 1 mol/1 L would do the trick.
Borek I literally just got your equation ahaha

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7 years ago
#15
(Original post by Branny101)
Borek I literally just got your equation ahaha
Using nonsensical concentrations. Highest concentration of anything at STP that I am aware of is that of a pure water - 55.5 moles per liter. You can't have 50 moles per mL - when you use such numbers you clearly show you have no idea what you are writing about.
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