[CHM 111] Nickel Quantification lab
Andrew J. Pounds
pounds_aj at mercer.edu
Fri Jun 14 20:18:33 EDT 2019
On 6/14/19 7:27 PM, wrote:
> For the nickel quantification lab, to calculated the concentration(M)
> for solutions 2-5, wouldn’t we need to first know the amount of
> solution 1 within each and calculate it that way. I am a little
> confused. Could you please lead me in the right direction for these
> calculations, thanks!
>
I have had this question come up a few times, so let me provide an example.
Let's say you weighed 3.9845 g of the nickel sulfate hexadrate.
To find the molarity of the first solution you find the moles of nickel
sulfate hexahydrate and then diluted it to 100 ml in the volumetric flask.
$M = \frac{\frac{3.9845}{262.9}}{0.100 L} = 0.1516\ M$
To make the second solution, you use 20 ml of this standard solution and
dilute it in a 50 ml volumetric flask. In this case, use the
$M_1V_1=M_2V_2$ equation in a rearranged format.
$M_2 = \frac{M_1V_1}{V_2} = \frac{(0.1516 M)(0.02000 L)}{(0.05000 L)} =
0.06064 M$
You then use the same process, i.e. $M_1V_1=M_2V_2$$, with varying
volumes to compute the molarities of solutions 3, 4, and 5. Be sure and
use the correct molarity of the initial solution when you do the
calculation for the fifth solution.
--
Andrew J. Pounds, Ph.D. (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Director of the Computational Science Program
Mercer University, Macon, GA 31207 (478) 301-5627
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