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Offline huskywolf

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HCl adjust ph of water problem....
« on: October 08, 2010, 01:06:37 PM »
Hi I have a problem here and I need some tips on how to solve it please....
Ground waters may contain hundreds of mg dm-3CO3-2 ions....

I have a 0.25 L sample of hard water (ground water) ,which is titrated with 0.073 M HCl , the ground water was found to require 0.035L of HCl to adjust the pH to 5.0


I need to calculate the estimated number of moles of CO3-2 in the 0.25L sample of hard water?
Please...

So far I have (0.035)(0.073) = 0.0026 M  HCl is used ,I don't know what to do next?

Offline DrCMS

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Re: HCl adjust ph of water problem....
« Reply #1 on: October 08, 2010, 01:38:13 PM »
Start with the reaction equation for the acid reacting with carbonate.

How many moles of acid react with each mole of carbonate?

How many moles of acid were used?

So how many moles of carbonate were in the sample?

Offline huskywolf

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Re: HCl adjust ph of water problem....
« Reply #2 on: October 08, 2010, 02:53:22 PM »
Hello,
Thanks for your reply....

I'm not sure which way the equation should go?

2HCl + CO32- = H2CO3 + Cl2+2 ?

For the number of moles of HCl i get seems wrong also:

n=M*v => N(HCl)=(0.0026)(0.035) =0.000091 mol?

Offline huskywolf

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Re: HCl adjust ph of water problem....
« Reply #3 on: October 08, 2010, 05:17:58 PM »
I got 0.000182 moles of carbonate ?

It looks correct since the the ratio of moles in the equation is 2:1


Is it correct please?

Offline huskywolf

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Re: HCl adjust ph of water problem....
« Reply #4 on: October 09, 2010, 04:16:41 AM »
Hi can someone tell me if I am correct please?

I need to be able to do this for tomorrow...
Please

Offline DrCMS

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Re: HCl adjust ph of water problem....
« Reply #5 on: October 09, 2010, 05:07:21 AM »
So far I have (0.035)(0.073) = 0.0026 M  HCl is used ,I don't know what to do next?

Not quite you've multiplied a molarity by a volume so the answer is moles not moles per litre.

For the number of moles of HCl i get seems wrong also:

n=M*v => N(HCl)=(0.0026)(0.035) =0.000091 mol?

No you have multiplied the no. of moles of HCl calculated above by the volume used.

Offline Borek

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Re: HCl adjust ph of water problem....
« Reply #6 on: October 09, 2010, 05:17:52 AM »
I'm not sure which way the equation should go?

2HCl + CO32- = H2CO3 + Cl2+2

What does it mean "balanced equation"?
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Offline huskywolf

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Re: HCl adjust ph of water problem....
« Reply #7 on: October 09, 2010, 08:20:48 AM »
So far I have (0.035)(0.073) = 0.0026 M  HCl is used ,I don't know what to do next?

Not quite you've multiplied a molarity by a volume so the answer is moles not moles per litre.

For the number of moles of HCl i get seems wrong also:

n=M*v => N(HCl)=(0.0026)(0.035) =0.000091 mol?

No you have multiplied the no. of moles of HCl calculated above by the volume used.

I see, thanks...

Now, I have the number of moles of HCl is 0.0026 , and the number of moles of carbonate is 0.0013 moles...
Is this correct now? Thanks

Offline huskywolf

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Re: HCl adjust ph of water problem....
« Reply #8 on: October 09, 2010, 08:23:15 AM »
I'm not sure which way the equation should go?

2HCl + CO32- = H2CO3 + Cl2+2

What does it mean "balanced equation"?

The mass on the left must = mass on the right....

Maybe it should look like this? 2HCl + H2CO3=> H2O + CO2+ 2HCl

Offline huskywolf

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Re: HCl adjust ph of water problem....
« Reply #9 on: October 09, 2010, 09:51:15 AM »
I'm not sure which way the equation should go?

2HCl + CO32- = H2CO3 + Cl2+2

What does it mean "balanced equation"?

The mass on the left must = mass on the right....

Maybe it should look like this? 2HCl + CO3=> H2O + CO2+
 2Cl-

Sorry ,, is the number of moles of Carbonate I calculated correct please?

Offline Borek

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Re: HCl adjust ph of water problem....
« Reply #10 on: October 09, 2010, 10:13:37 AM »
The mass on the left must = mass on the right....

That's only part. You also need identical charge on both sides.

Quote
Maybe it should look like this? 2HCl + H2CO3=> H2O + CO2+ 2HCl

You don't start with H2CO3, also if you have HCl on both sides and in the same amount, you should cancel them.

1.3 mmol looks OK, but IMHO you are far from being able to solve such questions.
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Offline huskywolf

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Re: HCl adjust ph of water problem....
« Reply #11 on: October 09, 2010, 10:29:57 AM »
The mass on the left must = mass on the right....

That's only part. You also need identical charge on both sides.

Quote
Maybe it should look like this? 2HCl + H2CO3=> H2O + CO2+ 2HCl

You don't start with H2CO3, also if you have HCl on both sides and in the same amount, you should cancel them.

1.3 mmol looks OK, but IMHO you are far from being able to solve such questions.

Thanks for the help,

I just need plenty of practice :)

Offline DrCMS

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Re: HCl adjust ph of water problem....
« Reply #12 on: October 09, 2010, 10:40:22 AM »
The balanced equation should be:

2HCl + CO32-  :rarrow: H2CO3 + 2Cl2-

also as  H2CO3  :rarrow: H2O + CO2


For a ground water sample it would most probably be:

2HCl + CaCO3  :rarrow: CaCl2 + H2O + CO2

Offline huskywolf

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Re: HCl adjust ph of water problem....
« Reply #13 on: October 09, 2010, 11:04:36 AM »
The balanced equation should be:

2HCl + CO32-  :rarrow: H2CO3 + 2Cl2-

also as  H2CO3  :rarrow: H2O + CO2


For a ground water sample it would most probably be:

2HCl + CaCO3  :rarrow: CaCl2 + H2O + CO2

Great, Thank you very much

Offline Borek

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Re: HCl adjust ph of water problem....
« Reply #14 on: October 09, 2010, 01:52:27 PM »
The balanced equation should be:

2HCl + CO32-  :rarrow: H2CO3 + 2Cl2-

You made the same mistake :D
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