Analyzing SO42- and PO43- in Aqueous Solution

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To determine the concentrations of SO42- and PO43- in an aqueous solution, a method involving titration with dilute HNO3 and NaOH is proposed. The initial step involves adding excess NaOH to convert the ions into their respective sodium salts, Na2SO4 and Na3PO4. Upon titration with HNO3, the strong acid (sulfuric acid from SO42-) and weak acid (phosphoric acid from PO43-) will react differently, allowing for differentiation based on their pKa values. The endpoint of the titration can be identified using an acid-base indicator, with the possibility of multiple endpoints for phosphoric acid. The discussion emphasizes the importance of understanding the reactions and the nature of the acids involved for accurate concentration determination.
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Homework Statement



An aqueous solution containing SO42- and PO43- is provided.
Using this solution,dil.HNO3 and dil NaOH only,outline a method to determine the concentration of SO42- and PO43- in the sample

Homework Equations



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The Attempt at a Solution



Well,I'm not really sure how to do this 'cos I'm not very sure of the reactions.Anyway this is what I think,
first, find the initial volume of the solution.next we have to find the no. of moles of each ion present, to find the concentration.
To find the no. of moles,
we add excess dilute NaOH to the sample,then we get 2 colourless solutions of Na2SO4 and Na3PO4.
To this solution we then add,dil nitric acid,Im thinking one would give a gas and one wouldn't.The one that gives out a gas contains sulphate ions,
3Na2SO4 + 6HNO3\rightarrow3SO2+6NaNO3+3H2O
Is the above equation correct?I know that a sulphite gives SO2 with an acid but does a sulphate also give the same products?
 
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Don't think about generating a gas... think about the pKa of the two acids (one with a sulfate counterion and the other with a phosphate). How can you use titration to solve this problem?
 
Sulphuric acid is a strong acid,whereas phosphuric acid is a relatively weak acid ,so... the Ka of sulphuric is much higher than that of phosphuric acid,which means the pKa is lesser for sulphuric than for phosphuric?
And umm...are the two acids that you're referring to, the ones that we get after adding nitric acid to their Na salts ?or is it only after adding dil.HNO3 to the (initial) given solution of sulphate and phosphate ions?
 
Yep. Now devise the experiment...
 
leena19 said:
Sulphuric acid is a strong acid,whereas phosphuric acid is a relatively weak acid ,so... the Ka of sulphuric is much higher than that of phosphuric acid,which means the pKa is lesser for sulphuric than for phosphuric?
And umm...are the two acids that you're referring to, the ones that we get after adding nitric acid to their Na salts ?or is it only after adding dil.HNO3 to the (initial) given solution of sulphate and phosphate ions?

chemisttree said:
Yep. Now devise the experiment...

I'm sorry,do you mean,we get phosphuric and sulphuric acid after we add HNO3 to the original solution?
SO42- +2 HNO3\rightarrowH2SO4 + 2NO3-

PO43- + 3HNO3\rightarrowH3PO4 + 3NO3-

Either way,
For the titration part,this is what I think,
Taking a known volume of the acid solution,( say 25cm3?) into a titration flask ,add few drops of phenolphthalein as an acid-base indicator,we titrate this using a known concentration of NaOH in the burette(say C).
but what happens now?
i know the reactions occur as follows,
NaOH + H2SO4 \rightarrow Na2SO4 + H2O

3NaOH + H3PO4 \rightarrow Na3PO4 + 3H2O
But what happens at the endpoint?
I know that when titrating with Na2CO3,there are 2 end points,with methyl orange as indicator,but does this apply here too ,for phosphuric acid?If so I don't know how and I can't help but feel like I 'm overthinking it.
Need some help here.please...
Thank you.
 
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