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Topic: pH of NaHSO4  (Read 19034 times)

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Offline alina0031

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pH of NaHSO4
« on: May 17, 2013, 12:49:36 AM »
To a chemistry experiment there is made an aqueous dissolution of NaHSO4. There is weighed 0,82 gram of the salt.  The salt is poured in a graduated flask where there is 100 mL of water .

1) What is the dissolution pH?

2) The is taken 10 mL of the dissolution with a pipette. The 10 mL of the dissolution is poured in a new graduated flask with 100 mL of water. What is the pH of the diluted dissolution?

(Sorry about my English i'm from Germany)

Offline Hunter2

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Re: pH of NaHSO4
« Reply #1 on: May 17, 2013, 01:34:56 AM »
What is your own attempt?  You need to calculate the moles of the solution. You need the pKA of hydrogensulfate and  the calculation formulas for weak acids.

Offline alina0031

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Re: pH of NaHSO4
« Reply #2 on: May 17, 2013, 02:21:58 AM »
What is your own attempt?  You need to calculate the moles of the solution. You need the pKA of hydrogensulfate and  the calculation formulas for weak acids.

I really don't know what to do.
I know that the molar mass of NAHSO4 is 120,0599 g/mol.
so the n = m / M => n = 0,82 g / 120,0599 =  0,00682992406291 mol

Then I say :
c = n / V => c = 0,00682992406291 mol / 0,1 L = 0,0682992406291 mol/L

Then I don't know what to do. And I'm not sure about the last part.



Offline Hunter2

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Re: pH of NaHSO4
« Reply #3 on: May 17, 2013, 02:48:56 AM »

Offline Borek

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Re: pH of NaHSO4
« Reply #4 on: May 17, 2013, 03:02:36 AM »
pH = ½(pKa - log [HA]0)

While your general advice about using Ka and weak acid formulas is OK, this is wrong.

Alina: you have a 0.0683M solution of a weak acid with Ka=10-1.9. Does it ring a bell?
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Offline alina0031

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Re: pH of NaHSO4
« Reply #5 on: May 17, 2013, 05:46:01 AM »
pH = ½(pKa - log [HA]0)

While your general advice about using Ka and weak acid formulas is OK, this is wrong.

Alina: you have a 0.0683M solution of a weak acid with Ka=10-1.9. Does it ring a bell?


But are my calculations correct?

Offline alina0031

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Re: pH of NaHSO4
« Reply #6 on: May 17, 2013, 05:51:21 AM »
- And is this correct:
pH = 1/2 *(pKs - logcs) => 0,5*(1,99-log(0,0683)) = 1,577

Offline Borek

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Re: pH of NaHSO4
« Reply #7 on: May 17, 2013, 06:18:10 AM »
You were OK up to the concentration of the hydrogensulfate. Unfortunately, the simplified formula Hunter suggested doesn't work here, it works only if the dissociation fraction of the weak acid is below 5%, which is not the case.

You should use ICE table.
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Offline alina0031

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Re: pH of NaHSO4
« Reply #8 on: May 17, 2013, 06:21:26 AM »
You were OK up to the concentration of the hydrogensulfate. Unfortunately, the simplified formula Hunter suggested doesn't work here, it works only if the dissociation fraction of the weak acid is below 5%, which is not the case.

You should use ICE table.

Thank you for helping me.
I am really lost. Can you write the solution.

Offline alina0031

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Re: pH of NaHSO4
« Reply #9 on: May 17, 2013, 06:39:24 AM »
Ups, I just found out how to solve it

Offline Hunter2

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Re: pH of NaHSO4
« Reply #10 on: May 17, 2013, 07:46:58 AM »
Then light the bulb please. Because I also get lost.

Erleuchte uns mal.

Offline Borek

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