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Topic: Concentration calculation  (Read 1377 times)

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Offline chehaza

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Concentration calculation
« on: April 25, 2020, 10:54:34 AM »
Hey im planing to do a metal complex experiment, If i have 5ml of 0.05M Ni(NO3)2+ and 5ml of ammonia how can i calculate the concentration of the product formed which is the nickel complex [Ni(NH3)6]2+. Do i simply just use C1*V1=C2*V2. I was possibly thinking of calculating the moles of the nickel nitrate and using that to find the moles of the complex, How could i take into account the moles of ammonia ?  any clarification would be great. Also i know this may seem like a straight forward question but ive been focused on organic chem for awhile.


Offline chenbeier

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Re: Concentration calculation
« Reply #1 on: April 25, 2020, 10:57:48 AM »
You have 0,05 M Ni2+, you want to make the hexaamincomplex, so how many moles ammonia do you need? This you have to calculate for 5 ml.

Offline AWK

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Re: Concentration calculation
« Reply #2 on: April 25, 2020, 11:02:25 AM »
Balance reaction. You should know the concentration of ammonia.
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Offline chehaza

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Re: Concentration calculation
« Reply #3 on: April 25, 2020, 11:04:00 AM »
I have 0.00025moles of Nickel nitrate, as amonnia has a 1:1 ratio i would also need 0.00025moles of ammonia? then I could calculate the concentration of ammonia but what then ?

Offline AWK

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Re: Concentration calculation
« Reply #4 on: April 25, 2020, 11:25:22 AM »
It is not possible to calculate stoichiometry of reactions without balancing.
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Offline chehaza

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Re: Concentration calculation
« Reply #5 on: April 25, 2020, 11:33:07 AM »
Thanks i just fixed up the equation i would get Ni(NO3)2 + 6NH3 = [Ni(NH3)6]2+. So therefore i can calculate the moles of nickel nitrate, i can then multiply by 6 to find the moles of ammonia. The moles then can be used to find concentration of ammonia but where would i go from there ? . Would i find the limiting reagent and then use the moles of that to find the moles of the nickel complex ?

Offline chenbeier

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Re: Concentration calculation
« Reply #6 on: April 25, 2020, 11:47:17 AM »
Correct. You know the mole of ammonia is 6 times. Now it depends what is the concentration of the ammonia what you will use. From this you can calculate the required volume.

Offline AWK

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Re: Concentration calculation
« Reply #7 on: April 25, 2020, 11:48:17 AM »
The ammonia complex with NI (II) is weak. If you want to practically obtain a complex with such stoichiometry, you need to add a significant excess of ammonia.
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Offline chehaza

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Re: Concentration calculation
« Reply #8 on: April 25, 2020, 12:03:32 PM »
Im still confused as to how i can calculate the moles of the nickel complex formed which would allow me to calculate the concentration. I have found the moles and concentration of ammonia but im still unsure where to go from there using either mole ratio would give me the same asnwer for the moles of the product.

Offline chehaza

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Re: Concentration calculation
« Reply #9 on: April 25, 2020, 12:22:24 PM »
Here is what I'm getting as the concentration. What's the point of me calculating the moles of ammonia if I'm gonna divide by 6 to get the moles of the nickel complex.
« Last Edit: April 25, 2020, 12:39:15 PM by chehaza »

Offline chenbeier

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Re: Concentration calculation
« Reply #10 on: April 26, 2020, 03:01:55 PM »
Correct, you need 0,0015 mol ammonia at least. As mentioned even more in excess. This ammount has to be in your 5 ml ammonia solution.

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