Titration calculation help!!!

icysniper

icysniper

icysniper

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Hi, I have a further chemistry exam tomorrow (GCSE) and am stuck on titration calculations. Here is the question below, if possible could you explain how you got your answer as I can't get my head around it atm :(

In a titration, a 25cm³ sample of nitric acid (HNO3) reacted exactly with 20cm³ of 0.40 mol/dm³ sodium hydroxide solution.

a) Calculate the number of moles of sodium hydroxide added
b) Write down the number of moles of NHO3 in the acid
c) Calculate the concentration of the nitric acid.
 
a) write out the reaction to get a balaced equation
b) the above allows you to get this,
c) (b)/ vol of Nictric acid (be careful with units!!)

At work so I can't do it exactly atm
 
feriso said:
a) write out the reaction to get a balaced equation
Click to expand...

balanced equation i get

HNO3 + NaOH = H20 + NaNO3

feriso said:
b) the above allows you to get this,
Click to expand...

so its a 1:1 ratio

heres the bit i don't get.....

for working out the moles of HNO3

moles = mass/rfm

25/(1+14+16+16+16 = 63)

= ~ 0.4 moles right??

so since its a 1:1 ratio, 0.4 moles of HNO3 reacted with 0.4 moles of NaOH??
 
From my admittedly 11 years out of date Chemistry knowledge it doesn't seem like your answering the question. At least what I think the answer to a) is has nothing to do with a balanced equation!
 
okay, there are 0.4 moles of NaOH in 1000cm³
so per 1cm³ there is

0.4/1000

= 0.0004 moles per cm³

x20 because theres 20cm³

= 0.008 moles of NaOH in 20cm³ and since its a 1:1 ratio there must be

0.008 moles of HNO3 in 25cm³

0.008/25

= 0.00032

x1000 to get it in mol/dm³

= 0.32 mol/dm³?????

EDIT:

anant_shah94 said:
are you doing OCR mate?
Click to expand...

no mate :) AQA
 
Last edited:
This is what I get for being too slow! Have something written out but it seems like feriso is guiding you through it. So as not to interfere i'll steer clear until you have a go at part c :)

EDIT: Yup - All looks fine to me :)

Hopefully Feriso will see if I’ve done something silly – which is far more likely that I’d like to admit; but here's what I had to give you:

a)
Points to remember:
-Conc is simply mol/volume which gives a clue to the calculation.
-Vol must be converted to dm3 (1000cm3 in 1dm3).
-Titration questions are always easier if you start off with a ‘balanced’ equation.
Moles = conc*vol
= 0.4*0.02dm3
=0.008 moles of sodium hydroxide titrated.

b) Need to get used to doing it yourself: but remind yourself why you’re doing this – the relationship is what’s important.
-1 mol HNO3 goes to give 1 mol NaOH. 1:1 relationship. (So we can skip ahead at this point, but its also ok to check yourself and your earlier answer by...)
-No big tricks here. At this level moles present is simply the mass / relative formula mass.
-I think at GCSE you’re also allowed to ignore density so that 1cm3 of solution is 1g! Useful.

= mass (or volume in cm3) / rfm
= 25 / 63[give or take]

= 0.3968 moles.
= 0.4 moles.

c) You had 0.008 moles present in 25cm3 of acid.
How many would be present in 1000cm3 of acid? (remember earlier we said that conc is given in mol/dm3.)

Couple easy steps should get you there: run it by us to check.
 
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icysniper said:
for working out the moles of HNO3

moles = mass/rfm

25/(1+14+16+16+16 = 63)

= ~ 0.4 moles right??

so since its a 1:1 ratio, 0.4 moles of HNO3 reacted with 0.4 moles of NaOH??
Click to expand...

Moles of nitric acid = moles of sodium hydroxide.

You shouldn't need to work out the molecular weight. You sure its not a solution of nitirc acid? Its really badly written and what the hell is mol/dm³? I assume thats M or MolL-1.
 
jms198 said:
Moles of nitric acid = moles of sodium hydroxide.

You shouldn't need to work out the molecular weight. You sure its not a solution of nitirc acid? Its really badly written and what the hell is mol/dm³? I assume thats M or MolL-1.
Click to expand...

hmm says in the question concentration of nitric acid so i assume you need to work out into mol/dm³??
 
jms198 said:
what the hell is mol/dm³? I assume thats M or MolL-1.
Click to expand...

This sort of thing really isn't helpful. Its a very commonly used terminology, especially at GCSE and this sort of nit-picking isn't necessary on the eve of his Exam. If you wish to help do so - go scaremonger elsewhere.
 
icysniper said:
okay, there are 0.4 moles of NaOH in 1000cm³
so per 1cm³ there is

0.4/1000

= 0.0004 moles per cm³

x20 because theres 20cm³

= 0.008 moles of NaOH in 20cm³ and since its a 1:1 ratio there must be

0.008 moles of HNO3 in 25cm³

0.008/25

= 0.00032

x1000 to get it in mol/dm³

= 0.32 mol/dm³?????
Click to expand...

a few may have missed this, but is this correct 0.32 mol/dm³ for the concentration of nitric acid??
 
jms198 said:
Moles of nitric acid = moles of sodium hydroxide.

You shouldn't need to work out the molecular weight. You sure its not a solution of nitirc acid? Its really badly written and what the hell is mol/dm³? I assume thats M or MolL-1.
Click to expand...

mol/dm3 is molar. dm3 = decimetre cubed, ie 10cmX10cmx10cm = 1L

a) its 0.4 M NaOH, so you have 20 mL * 0.4 M = 8 mmol NaOH
b) 8 mmol
c) 8 mmol in 25 mL, 8/25 = 0.32 M
 
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