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#1
please can someone list a method of finding titration calculations?
the CGP book and my-gcsescience.com do it different ways and i'm so confused
thank you!
0
3 years ago
#2
I learn the equation:
Moles = concentration X volume

ALWAYS remember If you're given a volume in cm*3 then divide it by 1000 to get it into dm*3

Firstly find out whatever you can from the information they provide, they usually give the concentration and volume so you can work out the moles

The moles is the same for both the reactants so substitute the moles into the equation to find the missing concentration/volume

Concentration is: moles divided by volume
Volume: moles divided my concentration

It's so hard to explain by typing 😂

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1
3 years ago
#3
(Original post by z1820)
I learn the equation:
Moles = concentration X volume

ALWAYS remember If you're given a volume in cm*3 then divide it by 1000 to get it into dm*3

Firstly find out whatever you can from the information they provide, they usually give the concentration and volume so you can work out the moles

The moles is the same for both the reactants so substitute the moles into the equation to find the missing concentration/volume

Concentration is: moles divided by volume
Volume: moles divided my concentration

It's so hard to explain by typing 😂

Posted from TSR Mobile
Great description! That is so helpful.

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0
3 years ago
#4
(Original post by jazz_xox_)
please can someone list a method of finding titration calculations?
the CGP book and my-gcsescience.com do it different ways and i'm so confused
thank you!
Usually you'll be given the volume and concentration of one of the reactants and the volume or concentration of the other.
Think of it as a neutralisation equation
25cm3 of 0.1 moldm3 HCI reacts with 22.5cm3 of NaOH
NaOH + HCI ----- NaCI + H2O
1) ensure equation is balanced.
2) I usually next write out a table, the two headings are the reactants: moles, volume and concentration need to be worked out.
Considering your always given the concentration and volume one reactant you can work out the number of moles. In the example: moles = conc X volume (0.1 X (25/1000) = 0.0025 moles
3) work out moles of other reactant. This is done by looking at the big numbers in front of the reactants 1HCI reacts with 1NaOH. Therefore there is a 1:1 ratio and the number of moles of HCI = moles of NaOH
Now you have NaOH moles (and given volume of 22cm3) you can work out concentration
0.0025/0.0022=1.13 moldm3 I think the working out is right but the method definitely is
Just remember your always given the volume and concentration of one. Multiply them to get moles. Use balanced equation to work out the number of moles of other reactant which you can then divide to work out either the concentration or volume
Hope this helped
2
#5
(Original post by z1820)
I learn the equation:
Moles = concentration X volume

ALWAYS remember If you're given a volume in cm*3 then divide it by 1000 to get it into dm*3

Firstly find out whatever you can from the information they provide, they usually give the concentration and volume so you can work out the moles

The moles is the same for both the reactants so substitute the moles into the equation to find the missing concentration/volume

Concentration is: moles divided by volume
Volume: moles divided my concentration

It's so hard to explain by typing 😂

Posted from TSR Mobile
thank you very much! this is how the my-gcsescience guy does it, but the text book confused me!
could you write down for the acid and alkali
concentration x volume = concentration x volume
because the moles are the same for each side? (this is if they want you to find a missing concentration or volume?)

thanks again
0
3 years ago
#6
(Original post by scphmapd)
Great description! That is so helpful.

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Just forgot to mention that the moles is that same if they both have the same number behind their formula

E.g.

2Na + 1Li

They don't have the same number of moles, Na has twice has many - say if you work out the moles of Sodium (Na) is 0.4 all you have to do is divide by 2 to get lithium (because it's half)

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0
3 years ago
#7
(Original post by jazz_xox_)
thank you very much! this is how the my-gcsescience guy does it, but the text book confused me!
could you write down for the acid and alkali
concentration x volume = concentration x volume
because the moles are the same for each side? (this is if they want you to find a missing concentration or volume?)

thanks again
SORRY! (edit)

forgot to mention that the moles is the same if they both have the same number behind their formula

E.g.

2Na + 1Li

They don't have the same number of moles, Na has twice has many - say if you work out the moles of Sodium (Na) is 0.4 all you have to do is divide by 2 to get lithium (because it's half)

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#8
(Original post by z1820)
SORRY! (edit)

forgot to mention that the moles is the same if they both have the same number behind their formula

E.g.

2Na + 1Li

They don't have the same number of moles, Na has twice has many - say if you work out the moles of Sodium (Na) is 0.4 all you have to do is divide by 2 to get lithium (because it's half)

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thank you ! would we be able to do a question together so i can be confident im doing it right?
0
3 years ago
#9
(Original post by jazz_xox_)
thank you ! would we be able to do a question together so i can be confident im doing it right?
Of course

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#10
(Original post by z1820)
Of course

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27.5 cm3 of 0.2 mol/dm3 hydrochloric acid is needed to titrate 25.0 cm3 of sodium hydroxide solution. What is the concentration of the sodium hydroxide solution?
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#11
(Original post by jazz_xox_)
27.5 cm3 of 0.2 mol/dm3 hydrochloric acid is needed to titrate 25.0 cm3 of sodium hydroxide solution. What is the concentration of the sodium hydroxide solution?
I got 0.22 mol/dm^3
0
3 years ago
#12
(Original post by jazz_xox_)
I got 0.22 mol/dm^3

Me too! (I hope it's right hahah)

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3 years ago
#13
(Original post by z1820)

Me too! (I hope it's right hahah)

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Sorry my units are wrong should be mol/dm*3

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#14
(Original post by z1820)
Sorry my units are wrong should be mol/dm*3

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Yay! thank you so much for your help, actually want one of these to come up now! good luck in your exam
0
3 years ago
#15
(Original post by jazz_xox_)
Yay! thank you so much for your help, actually want one of these to come up now! good luck in your exam
No problem, well done for learning so fast - they're really tricky! I think at least one question will come up on titration calculations & thank you, good luck to you too

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#16
(Original post by z1820)
No problem, well done for learning so fast - they're really tricky! I think at least one question will come up on titration calculations & thank you, good luck to you too

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I think where I was going wrong is I wasn't dividing by 1000 first
0
3 years ago
#17
(Original post by jazz_xox_)
I think where I was going wrong is I wasn't dividing by 1000 first
Oh I see ahaha
Are you feeling prepared for the exams?

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#18
(Original post by z1820)
Oh I see ahaha
Are you feeling prepared for the exams?

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Kind of, I'm nervous for C2 tbh because we have only had one (yes, one) lesson on C3 so I've been teaching myself that- but I got an A* in my C2 mock so fingers crossed! are you?

also someone just posted this question could you help? its fine if you dont want to its getting late
The equation for the reaction is

CaCO3(s) → CaO(s) + CO2(g)
Calculate the maximum mass of calcium oxide that can be obtained by
heating 25 tonnes of calcium carbonate.
(Relative atomic masses: C = 12, O = 16, Ca = 40)

[The answer is 25 (tonnes) gives 56x25 (tonnes)]
0
3 years ago
#19
(Original post by jazz_xox_)
Kind of, I'm nervous for C2 tbh because we have only had one (yes, one) lesson on C3 so I've been teaching myself that- but I got an A* in my C2 mock so fingers crossed! are you?

also someone just posted this question could you help? its fine if you dont want to its getting late
The equation for the reaction is

CaCO3(s) → CaO(s) + CO2(g)
Calculate the maximum mass of calcium oxide that can be obtained by
heating 25 tonnes of calcium carbonate.
(Relative atomic masses: C = 12, O = 16, Ca = 40)

[The answer is 25 (tonnes) gives 56x25 (tonnes)]
Hmmm there's 2 ways but:

25 tonnes -------> ?
Mr: (adding all the relative atomic masses)
CaCo3----->CaO
100 ------> 56

100 X 0.56 = 56
So 25 X 0.56 = 14

0.56 X 25 = 14 tonnes ?

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#20
(Original post by z1820)
Hmmm there's 2 ways but:

25 tonnes -------> ?
Mr: (adding all the relative atomic masses)
CaCo3----->CaO
100 ------> 56

100 X 0.56 = 56
So 25 X 0.56 = 14

0.56 X 25 = 14 tonnes ?

Posted from TSR Mobile
exactly what I got. the person in another forum was arguing for it to be 56 because its in tonnes? i didn't understand
you are actually a life saver for helping me!! i'm getting off TSR now so good luck tomorrow i'm sure you'll do amazing
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