Titration calculation help!!!

icysniper · 25 May 2010 at 13:28

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.

icysniper · 25 May 2010 at 13:41

feriso said:
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

thanks

will get to work on it

icysniper · 25 May 2010 at 13:51

feriso said:
a) write out the reaction to get a balaced equation

balanced equation i get

HNO3 + NaOH = H20 + NaNO3

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

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??

icysniper · 25 May 2010 at 14:04

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?

no mate

AQA

icysniper · 25 May 2010 at 14:23

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.

hmm says in the question concentration of nitric acid so i assume you need to work out into mol/dm³??

icysniper · 25 May 2010 at 14:28

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?????

a few may have missed this, but is this correct 0.32 mol/dm³ for the concentration of nitric acid??

icysniper · 25 May 2010 at 14:40

eLbot said:
Yup

wooooo!!!! yay now time to do some more which i have printed off -_- if i need any help, ill just post some more

lol which is probably a 100% chance

icysniper · 25 May 2010 at 15:31

heres another then:

25cm³ of sodium carbonate reacted with 20cm³ of 0.2mol/dm³ of nitric acid

Na2CO3 + 2NaNO3 = 2NaNO3 + CO2 + H20

a) Calculate the number of moles of nitric acid in 20cm³.
b) Calculate the number of moles of sodium carbonate in 25cm³.
c) Calculate the concentration of sodium carbonate in mol/dm³

icysniper · 25 May 2010 at 15:40

okay, my answer for a) is 0.63 moles of nitric acid

20/63

x2

= 0.63

since its 1:2, there must be half the moles for sodium carbonate

b) 0.63/2= 0.315

then for c) i did:

0.2/1000

x20

=0.004

1:2

so half 0.004 to get 0.002

0.002/25

x1000

= 0.08mol/dm³, is this correct???? :S

icysniper · 25 May 2010 at 15:41

jms198 said:
Equation is wrong Na2CO3 is sodium carbonate.

woops that is my mistype the equation should be

Na2CO3 + 2HNO3 = 2NaNO3 + CO2 + H20

icysniper · 25 May 2010 at 16:45

thanks everyone for their help, has been very useful and as of now i understand titration calculations, thank you OcUK for your amazingness (not a word) for helping me out!!!