Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: chehaza on April 25, 2020, 10:54:34 AM
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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.
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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.
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Balance reaction. You should know the concentration of ammonia.
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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 ?
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It is not possible to calculate stoichiometry of reactions without balancing.
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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 ?
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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.
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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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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.
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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.
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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.