please Help with these concentration calculations! It\'s for my lab and i\'m not

ID: 893113 • Letter: P

Question

please Help with these concentration calculations! It's for my lab and i'm not sure how to do it. Need help ASAP! Will reward points to best answer! Thanks!
Abbreviated Procedure 1. sample preparation-total volume of 10 mL for each sample is more than enough a) two stock solutions will be provided (nothing goes into stock solutions, pour a small amount into a beaker for your use): 2.00 x 10' M quinine sulfate dihydrate in - 10 M H,so. A+ b) make 5 samples with different concentrations of CT in the range of 00-004 M- exact c) concentration of quinine sulfate dihydrate should be near 1 00 × 10" M and needs to be d) concentration of H:SO4 should be- 0.5 M 0.150 M KCI value is unimportant but value should be known to 3 places the same for all samples

Explanation / Answer

Hello, this is very simple, but you have to solve it step by step. First we need the moles of every compound.

Remember that:

M = n / V (L)

0.010 L = 10 ml

With that in mind let’s solve:

b) For the first sample don’t take any KCl at all, so you have 0.0 M of Cl-

For 0.01 M of Cl- we need:

moles of Cl- = M * V (L) = 0.01 M * 0.01 L = 0.0001 moles

With the amount of moles needed we can get the volume you’ll extract from the beaker with 0.150 KCl.

V = 0.0001 moles / 0.150 M = 0.0007 L = 0.7 ml

So, let’s put this very clear, you have a flask with a little amount of KCl 0.150 M, so accordingly to our math, you’ll extract 0.7 ml from there and take it to the second sample.

For the rest KCl we do exactly the same steps:

KCl (0.150 M)

1 (0.0 M)

2 (0.01M)

3 (0.02 M)

4 (0.03 M)

5 (0.04 M)

Volume to extract from the beaker

0.7 ml

1.3 ml

2.0 ml

2.7 ml

Try to have a beaker with 10 ml of KCl then.

c) In order to the concentration of quinine is the same in all the samples we do the same procedure of above. This time the volume will vary because from sample 2 to 5 you don’t have 10 ml available, but 10 ml - ml of KCl. Again, we need to do a table. I’ll do the first and the second example:

Sample 1:

moles of quinine = 1*10-5 M * 0.01 L = 1*10-7 moles

With the amount of moles needed we can get the volume you’ll extract from the beaker with 2*10-5 M of quinine.

Volume = 1*10-7 moles / 2*10-5 M = 0.005 L = 5 ml

So, you take 5 ml from the beaker.

Sample 2:

moles of quinine = 1*10-5 M * (0.01 - 0.0007 L) = 9.3*10-8 moles

volume = 9.3*10-8 moles / 2*10-5 M = 0.00465 = 4.7 ml

Quinine 2*10-5 M

sample 1

sample 2

sample 3

sample 4

sample 5

Volume needed

5 ml

4.7 ml

4.35 ml

4 ml

3.65 ml

So far, you have:

Total volume (parcial)

5 ml

5.4 ml

5.65 ml

6 ml

6.33

d) To complete each solution, you take H2SO4 and reach the 10 ml mark. Let’s calculate what would be the concentration of H2SO4 on each sample:

sample 1:

volume need to complete the mark: 5 ml

moles of H2SO4 = 1 M * 0.005 L = 0.005 moles

[H2SO4] = 0.005 / 0.01 L = 0.5 M

sample 2:

volume need to complete the mark: 10 ml - 5.4 = 4.6 ml

moles of H2SO4 = 1 M * 0.0046 L = 0.0046 moles

[H2SO4] = 0.0046 / 0.01 L = 0.46 M

sample 3:

volume need to complete the mark: 10 ml - 5.65 = 4.35 ml = 4.4 ml

moles of H2SO4 = 1 M * 0.0044 L = 0.0044 moles

[H2SO4] = 0.0046 / 0.01 L = 0.44 M

sample 4:

volume need to complete the mark: 10 ml - 6 = 4 ml

moles of H2SO4 = 1 M * 0.004 L = 0.004 moles

[H2SO4] = 0.004 / 0.01 L = 0.4 M

sample 5:

volume need to complete the mark: 10 ml - 6.33 = 3.7 ml

moles of H2SO4 = 1 M * 0.0037 L = 0.0037 moles

[H2SO4] = 0.0037 / 0.01 L = 0.37 M almost 0.4 M