x Turn on thread page Beta
 You are Here: Home

# Titration Calculations help watch

1. Having done a question for my homework for numerous reasons i believe my answer to be incorrect, the questions and my response are as follows, and would greatly appreciate any help as to where i've gone wrong (as my final answer would be tellurium which seems rather unlikely and odd, as it would have to be an unusual isotope of tellurium and tellurium doesn't seem like a common answer)

A solution is made by dissolving 5.00g of Sodium hydroxide to make 1 dm3 of solution. A solution of and acid H2X is made by dissolving 7.66g to make 1 dm3 of solution. 25.0cm3 of the sodium hydroxide solution is neutralised by 20.0cm3 of the acid solution.

a)Write an equation for the reaction
2NaOH + H2X --> Na2X + 2H2O

b)Calculate the concentration of the sodium hydroxide in moldm^-3
NaOH conc = moles/vol
moles=mass/Mr
moles=5/40
1/8 moles, 1dm^3, thus 0.125moldm^-3

c)Calculate the amount in moles of sodium hydroxide in 25.0cm^3
(0.125/1000)*25=0.003125 moles

d)Calculate the amount of moles of H2X in 20.0cm^3 of solution.
2:1 ratio of NaOH to H2X, therefore 1/16 moles of H2X, it's dissolved to make 1dm^3 of solution, therefore 1/16moldm^3 concentration.
1/16moldm^-3 * 0.020dm^3 = 0.00125 moles

e)Calculate the concentration of H2X in moldm^-3
(see above)

f)Calculate the mass of the anion, X2-
Mass/Moles=Mr
7.66/(1/16)=122.56
122.56-2=Ar of X
Ar of X=120.56

g)Suggest an identity for the anion X2-
Tellurium has an isotope with an Ar of 120, and as it is a group 6 element would form a 2- anion.

Any help as to any blunders i've made would be much appreciated, thank you
2. (Original post by fw431)
Having done a question for my homework for numerous reasons i believe my answer to be incorrect, the questions and my response are as follows, and would greatly appreciate any help as to where i've gone wrong (as my final answer would be tellurium which seems rather unlikely and odd, as it would have to be an unusual isotope of tellurium and tellurium doesn't seem like a common answer)

A solution is made by dissolving 5.00g of Sodium hydroxide to make 1 dm3 of solution. A solution of and acid H2X is made by dissolving 7.66g to make 1 dm3 of solution. 25.0cm3 of the sodium hydroxide solution is neutralised by 20.0cm3 of the acid solution.

a)Write an equation for the reaction
2NaOH + H2X --> Na2X + 2H2O

b)Calculate the concentration of the sodium hydroxide in moldm^-3
NaOH conc = moles/vol
moles=mass/Mr
moles=5/40
1/8 moles, 1dm^3, thus 0.125moldm^-3

c)Calculate the amount in moles of sodium hydroxide in 25.0cm^3
(0.125/1000)*25=0.003125 moles

d)Calculate the amount of moles of H2X in 20.0cm^3 of solution.
2:1 ratio of NaOH to H2X, therefore 1/16 moles of H2X, it's dissolved to make 1dm^3 of solution, therefore 1/16moldm^3 concentration.
1/16moldm^-3 * 0.020dm^3 = 0.00125 moles

e)Calculate the concentration of H2X in moldm^-3
(see above)

f)Calculate the mass of the anion, X2-
Mass/Moles=Mr
7.66/(1/16)=122.56
122.56-2=Ar of X
Ar of X=120.56

g)Suggest an identity for the anion X2-
Tellurium has an isotope with an Ar of 120, and as it is a group 6 element would form a 2- anion.

Any help as to any blunders i've made would be much appreciated, thank you

D is where you get messy.
There are 3.125x10-3 moles of NaOH.
Because of the 2:1 ratio, the moles of H2X that should have reacted with this is therefore 1.5625x10-3 moles.

With that in mind, you can now calculate concentration, and subsequently the mass.
Spoiler:
Show
I don't particularly think the anion necessarily has to be an whole element. Once you get to the answer I think you might realise what I mean.
3. (Original post by RMNDK)
D is where you get messy.
There are 3.125x10-3 moles of NaOH.
Because of the 2:1 ratio, the moles of H2X that should have reacted with this is therefore 1.5625x10-3 moles.

With that in mind, you can now calculate concentration, and subsequently the mass.
Spoiler:
Show
I don't particularly think the anion necessarily has to be an whole element. Once you get to the answer I think you might realise what I mean.
Thank you so much, having followed through with the readjustments i see what you mean, as i ended up concluding that X is SO4

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: November 12, 2015
Today on TSR

### Negatives of studying at Oxbridge

What are the downsides?

### Grade 9 in GCSE English - AMA

Poll

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE