Question

Five standard solutions of HBr are prepared by serial dilution in which, at each stage, 10.00mL are diluted to 150.00mL. Given that the concentration of the most dilute solution is 2.37x10^-6 M, determine the concentration of the original HBr stock solution

Answer 1

I got .00004 M but it doesn't seem right.

Answer 2

What are your steps?

Answer 3

I know I have to 2.37x10^-6 x 150 mL = M x 10 mL for about 4 of them.

Answer 4

I'm not sure how to solve this.

Answer 5

mk

Answer 6

Thanks for looking.

Answer 7

@sweetburger

Answer 8

I could post how someone explained it and see if it makes sense to you. If you would like.

Answer 9

If you make your example by starting let's say with 0.002 mol in 0.01 L solution (which is 10 mL) and keep on diluting it to 0.15L (which is 150 mL) and calculate the final M, you'll see that it is decreasing in molarity by a factor of 15. Which makes sense because 0.15L is 15 times greater that 0.01. If you take the molarity of the diluted solution multiplied by 15 you'll get the molarity of the previous solution. So I think to get back to you're original concentration it is final concentration multiplied by 15 raised to the 5 -1

\[final M (15)^{5 -1}\]

You can verify if your answer is correct by diluting 5 timesto 150 mL (0.15L)

Answer 10

I get .11998125 when I do it that way.

Answer 11

So take 0.112 M calculate the moles of 0.01L and then dilute (that mole) to 0.15L and repeat for a total of 5 and see if you get 2.37x10^-6.

Answer 12

Where is the .112 M coming from?

Answer 13

I get .11998125 rounded

Answer 14

I get that too.

Answer 15

Why wouldn't it be .12 M?

Answer 16

You are correct. Verify if that is the correct answer.

Answer 17

2.73x10^-6 *10.00/150.00 = 1.82 x 10^-7

Answer 18

1.82x10^-7 *10.00/150.00 = 1.21x10^-8

Answer 19

1st concentration:

0.12 M x 0.01L = 1.2x10^-3 mole

2nd concentration:

(1.2x10^-3)/0.15 = 8 x 10^-3 M

3rd concentration

8 x 10^-3 x 0.01 = 8 x 10^-5 mol

(8 x 10^-5) / 0.15 = 5.3333 x 10^-4 M

4th concentration

5.3333 x 10^-4 x 0.01 = 5.3333 x 10^-6 mol

(5.3333 x 10^-6) / 0.15 = 3.5555 x 10^-5 M

5th concentration

3.5555 x 10^-5 x 0.01 = 3.5555 x 10^-7

(3.5555 x 10^-7) / 0.15 = ?

Answer 20

2.370 x 10^-6

Answer 21

Is .12M the right amount of sig figs?

Answer 22

It should probably be 0.120 M. Do you submit the answer online?

Answer 23

Yes.

Answer 24

Is it the correct answer?

Answer 25

It counted it wrong but it makes sense though. If you can plug it back in like that.

Answer 26

Is this AP chemistry or college chemistry? Can you email your professor and show the work.

I think we made a slight mistake. The question might be, a stock solution (that was already prepared) was diluted 5 times. So it would be raised to the 5th power (not 5 -1).

How many tries do you have?

Answer 27

That was my last try. It's college.

Answer 28

Can you let me know the correct answer.

Answer 29

Is it a regular chem class or honors?

Answer 30

If I ever find out. It doesn't tell you the right answer afterwards for some reason.

Answer 31

It's a regular chem class.

Answer 32

I think you should email your professor, you may get partial credit.

Answer 33

I will. Thanks

Answer 34

You can use M\(_1\)V\(_1\) = M\(_2\)V\(_2\) to solve.
Essentially, you're going to be doing this a few times since this was a serial dilution that you did.

Or, you can simply figure out your dilution factor(s) (DF) and go from there using DF = \(\sf \frac{C_1}{C_2}\)