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November 01, 2025, 12:20:49 pm

Author Topic: Dilution/concentration Question  (Read 865 times)  Share 

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Snorlax

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Dilution/concentration Question
« on: August 06, 2013, 08:03:05 pm »
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What are the final concentrations of sodium, potassium and sulfate ions when 200cm3 of 2.00M sodium sulfate solution are mixed with 300cm3 of 1.00M potassium sulfate solution?

I understand what the question wants, but just can't figure out how to solve it.
Help will be awesome!
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lzxnl

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Re: Dilution/concentration Question
« Reply #1 on: August 06, 2013, 08:31:18 pm »
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Work out the number of moles of sodium and potassium ions. Also, work out the number of sulfate ions in each solution and then add these to get the total amount of sulfate. Then, remembering that the volume is now 500mL, work out the concentrations.
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RKTR

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Re: Dilution/concentration Question
« Reply #2 on: August 06, 2013, 08:33:03 pm »
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What are the final concentrations of sodium, potassium and sulfate ions when 200cm3 of 2.00M sodium sulfate solution are mixed with 300cm3 of 1.00M potassium sulfate solution?

I understand what the question wants, but just can't figure out how to solve it.
Help will be awesome!
Na2SO4--> 2Na+  +  SO4
K2SO4--> 2K+  + SO4

no.of mol of sodium sulfate= 0.200 x 2.00 = 0.400mol
no. of mol of potassium sulfate=0.300x1.00=0.300mol

1 mol sodium sulfate gives 2 mol of sodium ions
so no. of mol of Na ions =0.800 mol
c=0.800/0.500=1.60 M

no.of mol of K ions =0.600mol
c=0.600/0.500=1.20M

no.of mol of SO4 ions= 0.400+0.300=0.700mol
c=0.700/0.500=1.40M
 
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Snorlax

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Re: Dilution/concentration Question
« Reply #3 on: August 06, 2013, 08:50:19 pm »
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Na2SO4--> 2Na+  +  SO4
K2SO4--> 2K+  + SO4

no.of mol of sodium sulfate= 0.200 x 2.00 = 0.400mol
no. of mol of potassium sulfate=0.300x1.00=0.300mol

1 mol sodium sulfate gives 2 mol of sodium ions
so no. of mol of Na ions =0.800 mol
c=0.800/0.500=1.60 M

no.of mol of K ions =0.600mol
c=0.600/0.500=1.20M

no.of mol of SO4 ions= 0.400+0.300=0.700mol
c=0.700/0.500=1.40M

Thanks a ton!
Easy to follow along :)
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RKTR

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Re: Dilution/concentration Question
« Reply #4 on: August 06, 2013, 08:54:27 pm »
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Thanks a ton!
Easy to follow along :)
you're welcome  :)
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