AS level chemistry question.. titrations HELP PLEASE

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beccajonesy
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Hi any help with the quesstion below is much appricated.

5.125g of washing soda crystals are dissolved and made up to 250cm3 of solution. A 25cm3 portion required 35.8cm3 of 0.0500 mol dm-3 sulphuric acid for neutralisation. Caculate the % of sodium carbonate in the crystals.

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Revolution is my Name
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Work out the number of moles of Sulphuric Acid in the solution (using the equation Moles= Volume x Concentration, and remembering to put the concentration in dm^3). Then write a balanced equation, from which it should become apparent how many moles of Sodium Carbonate are required to react with this number of moles. Then multiply this by 10 to get the number of moles in the full 250cm^3. Then use the equation Mass= Molar Mass x Number of moles, which will give you the mass of Sodium Carbonate in the soda crystals. Then divide this by 5.125, and multiply by 100, to get the Sodium Carbonate in the crystals.
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beccajonesy
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(Original post by Revolution is my Name)
Work out the number of moles of Sulphuric Acid in the solution (using the equation Moles= Volume x Concentration, and remembering to put the concentration in dm^3). Then write a balanced equation, from which it should become apparent how many moles of Sodium Carbonate are required to react with this number of moles. Then multiply this by 10 to get the number of moles in the full 250cm^3. Then use the equation Mass= Molar Mass x Number of moles, which will give you the mass of Sodium Carbonate in the soda crystals. Then divide this by 5.125, and multiply by 100, to get the Sodium Carbonate in the crystals.
Any idea what the balanced equation is ??
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RobG93
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Sodium Carbonate + Sulphuric Acid -----> Sodium Sulphate + Carbonate Acid

Na2CO3 + H2SO4 -----> Na2SO4 + H2CO3
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Nan8903
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why multiplying 10 get your no. of moles in full 250cm³ solution
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Pigster
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(Original post by Nan8903)
why multiplying 10 get your no. of moles in full 250cm³ solution
Imagine the original 250 cm3 solution had 10 mol in it and you took out 25 cm3 of that solution, it would contain 1 mol.

Now imagine you took 25 cm3 out of a 250 cm3 solution and you found that the 25 cm3 had I don't know 0.0125 mol in. Surely the 250 cm3 would contain 0.125 mol?!?
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DavidX123456
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(Original post by Pigster)
Imagine the original 250 cm3 solution had 10 mol in it and you took out 25 cm3 of that solution, it would contain 1 mol.

Now imagine you took 25 cm3 out of a 250 cm3 solution and you found that the 25 cm3 had I don't know 0.0125 mol in. Surely the 250 cm3 would contain 0.125 mol?!?
Three words : 11 years ago
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Pigster
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(Original post by DavidX123456)
Three words : 11 years ago
11 isn't a word and the thread is only ten years old?

I was replying to Nan8903, who was asking a Q about a thread they have somehow found. Perhaps they were doing a past paper with this Q and googled "5.125 g of washing soda crystals" and found this thread.

I usually say something along the lines of "let the dead rest", but was too sleepy this morning to be sarcastic. When people reply to threads giving solutions to an OP from many many years ago, then they get both barrels. My favorite is when people give solutions that are wrong to threads that already have the correct solution posted.
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EierVonSatan
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(Original post by Pigster)
11 isn't a word and the thread is only ten years old?

I was replying to Nan8903, who was asking a Q about a thread they have somehow found. Perhaps they were doing a past paper with this Q and googled "5.125 g of washing soda crystals" and found this thread.

I usually say something along the lines of "let the dead rest", but was too sleepy this morning to be sarcastic. When people reply to threads giving solutions to an OP from many many years ago, then they get both barrels. My favorite is when people give solutions that are wrong to threads that already have the correct solution posted.
Once upon a time, threads would lock after no posts for a number of years. Such as https://www.thestudentroom.co.uk/showthread.php?t=65926 (if you alter the number at the end of the url you'll see that all similar threads are locked - so it wasn't a manual process).

For me the more frustrating ones is where a solution is given in great detail and then someone 10 years later asks a question about it but they haven't bothered to read the thread properly :mad:
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