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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: ziagemini on March 07, 2012, 07:33:10 AM

Hello All,
I need some help regarding the concentration calculation. I have 2 solution i.e. 0.1M NH4CL and 0.1M CH3COOH. Now i take 25ml of each solution and mix then. Now the total solution is 50ml. How can i calculate what amount of acetate and ammonium present in the solution.
Hope i get some good response. Thanks for the help.
Regards!

How much total acetate and total ammonium were present in your original solutions?

For simple dilutions C1*V1=C2*V2 is quick and sure. Where C1 is equal to the initial concentration and C2 is the final concentration. V1 is initial volume, V2 is final volume.

Hi,
I prepared standard solutions from stock solutions. That is:
1. 0.1M Ch3COOH= by adding 5.74ml of 99% pure acid into 994.2ml distilled water
2. 0.1M NH4CL = by adding 5.349 gram powdered NH4CL into 1 L distilled water.
From these two i took 25 ml of each solution and mixed it. So total solution was 50ml (25ml Ch3COOh + 25ml NH4CL). And then i titrate the mixture with 0.1M NaOH. I would like to calculate how much acetate and ammonium were present in 50 ml soltion before titration.
Hope i explained it correctly. Thanks for the help.
Regards!

yes, you explained correctly what you did. What I am asking you to determine now is, how much acetate (moles) was present in the 25 mL of solution you used? How much ammonium (moles) was present in the 25 mL of that solution that you used? (hint  what is the definition of M in your 0.1M solutions?)

Hello,
0.1M means 0.1 moles/Litre. so if i go with your hint that means 0.1*0.025 moles of acetate and ammonium were present in 25ml sol. and if i convert moles into gram then 0.150g (0.0025*60.05) Acetate and 0.045g(0.0025*18.03) of ammonium were present.
Is that so?
Regards!

Hello people,
Is my above calculations right? I don't think so. One of my friend said that each one is of 0.05M (mol/L) concentration. But he don't tell how he calculate that. Can anybody please tell me how to calculate that: Please. Thanks for the help and time.
Regards!
Zia

They ionize 1:1, so no added difficulty there.
You are taking a 0.1M of each and halving it's respective concentration (1:1 v/v mix of the two solutions). 0.1M/2 =0.05M.

Yes its ok. So what if i take 10ml of acetic acid and 40 ml of Nh4CL? Or if i mix three or four solutions. My purpose here is to know the way to calculate myself. Please tell me how to calculate the.
Lets say 1. 10ml of 0.1M Ch3COOh and 40 ml of 0.1M Nh4CL is mixed?
2. What if i mix 10ml o.1M Ch3COOH 25ml of 0.1M Na2CO3 and 15ml Nh4Cl?
How will i calculate in those situations.
Thanks

http://www.chembuddy.com/?left=concentration&right=dilutionmixing

If you are looking for your amounts in terms of concentration, Jasim already gave you the simple form:
For simple dilutions C1*V1=C2*V2 is quick and sure. Where C1 is equal to the initial concentration and C2 is the final concentration. V1 is initial volume, V2 is final volume.
You know the initial concentration and volume, you know the final volume, solve for final concentration.

Ok Ok OK. Now i got it i think ???
What if i mix 10ml o.1M Ch3COOH 25ml of 0.1M Na2CO3 and 15ml Nh4Cl?
So using the Jasim formula C1V1=C2V2 i calculate the concentration of each with respect to total volume i.e. C(acetic acid)=0.02M; C(Na2CO3)=0.05M and C(Nh4CL)=0.03M.
Thanks a lot. Thanks. Atlast this thing got in my head.
Thank you all of you for your help.
Regards!

Note  acetic acid reacts with sodium carbonate!

Hello,
Can somebody please elaborate what does that means ''Note  acetic acid reacts with sodium carbonate''. As, i am planning to prepare a solution of the above and then titrate it with HCL.
The idea behind that is there will be 2 buffers in the solution i.e. Carbonate and Acetic acid. I will have a buffer action at around pH 6.5 and then around 4.7. Am i doing something wrong? What other chemicals i can use to have carbonate buffer. I thought about CaCO3 but its is not soluble in water.
Thanks for the kind advice.
Regards!

You buffer contains NaHCO_{3} and Na_{2}CO_{3}. Moreover NH_{4}Cl change the ionic strength  this couse change of pH by ~0.05 pH unit.