ps1265A
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Sketch pH curves for following titration, 20cm3 0.10mol dm-3 NH3 against 0.1mol dm-3 HCl

The answer is:
http://postimg.org/image/9prhrps7b/


How do I go about:
1) Getting initial pH
2) End-point (I think I know how to - equate the moles)
3) Final pH

This is a weak base-strong acid, and the graph doesn't show this does it?
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charco
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(Original post by ps1265A)
Sketch pH curves for following titration, 20cm3 0.10mol dm-3 NH3 against 0.1mol dm-3 HCl

The answer is:
http://postimg.org/image/9prhrps7b/


How do I go about:
1) Getting initial pH
2) End-point (I think I know how to - equate the moles)
3) Final pH

This is a weak base-strong acid, and the graph doesn't show this does it?
Initial pH is from ka and concentration. It's a calculation, but you can work on the rule of thumb for ammonia that 0.1M will be about pH11.
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ps1265A
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(Original post by charco)
Initial pH is from ka and concentration. It's a calculation, but you can work on the rule of thumb for ammonia that 0.1M will be about pH11.
Thanks! Is my method for calculating equivalence point correct?
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Borek
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General approach to the titration curve:

http://www.titrations.info/titration-curve-calculation

Acid base titration curve:

http://www.titrations.info/acid-base...ve-calculation

You start with a solution of a weak base, so the simplest approach is to calculate pOH and convert it to PH.

End point - is just a pH of a salt solution, NH4+.

After that, pH is almost solely described by the dilution of the excess acid added.
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ps1265A
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(Original post by Borek)
General approach to the titration curve:

http://www.titrations.info/titration-curve-calculation

Acid base titration curve:

http://www.titrations.info/acid-base...ve-calculation

You start with a solution of a weak base, so the simplest approach is to calculate pOH and convert it to PH.

End point - is just a pH of a salt solution, NH4+.

After that, pH is almost solely described by the dilution of the excess acid added.
Thanks
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ps1265A
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(Original post by charco)
Initial pH is from ka and concentration. It's a calculation, but you can work on the rule of thumb for ammonia that 0.1M will be about pH11.
Why can I not calculate the concentration of the weak acid by using the pH when no NaOH is added?
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ps1265A
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(Original post by charco)
Initial pH is from ka and concentration. It's a calculation, but you can work on the rule of thumb for ammonia that 0.1M will be about pH11.
Why does the reaction between Na2CO3 and HCl have 2 equivalence points? Is it become 2 moles of HCl react? I thought for it to be two equivalence points, there would have to be a diprotic acid presents of a base which accepts 2 protons???
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charco
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(Original post by ps1265A)
Why can I not calculate the concentration of the weak acid by using the pH when no NaOH is added?
If you know the ka you can ...
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charco
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(Original post by ps1265A)
Why does the reaction between Na2CO3 and HCl have 2 equivalence points? Is it become 2 moles of HCl react? I thought for it to be two equivalence points, there would have to be a diprotic acid presents of a base which accepts 2 protons???
The carbonate ion gets neutralised in two steps:

CO32- + H+ --> HCO3- (no gas produced)

HCO3- + H+ --> CO2 + H2O
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ps1265A
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(Original post by charco)
If you know the ka you can ...
Can I not just do 10^-1.6 from the graph?
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ps1265A
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(Original post by charco)
The carbonate ion gets neutralised in two steps:

CO32- + H+ --> HCO3- (no gas produced)

HCO3- + H+ --> CO2 + H2O
CO2 and H20 are produced because there's a carbonate involved, thanks!

Is that always the case, first reaction is no gas and second reaction is gas? If we look back on my previous thread http://www.thestudentroom.co.uk/show....php?t=3042477 the answer to part e) is 15cm3 (which is after the first reaction and before the second)
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