Serial Dilutions - AS Biology

Hi, I was wondering whether there was a formula to calculate the end concentrations of serial dilutions and how I would apply this to questions.

E.g. 1cm^3 of 2M hydrochloric acid was added to 9cm^3 of distilled water. The mixture was inverted, then 1cm^3 of the mixture was added to a further 9cm^3 of distilled water. What was the final concentration?

Thanks in advance. :smile:

C1*v1=c2*v2

0.02moldm^-3

OP

Original post by Jsisnxjdnd

C1*v1=c2*v2

Ah thank you, but how would you apply this to this question in particular?

Reply 4

anona1255

OP

5

Original post by aytuiq

0.02moldm^-3

Thank you

Reply 5

aytuiq

15

No problem !

Good luck with your exam !

Reply 6

aytuiq

15

Explanation to answer :

1cm^3 2M HCl added to 9cm^3 of water means there is a final volume of 10cm^3 with a dilution of 1:10 and 2M/10=0.2M,
This is repeated with 9cm^3 again, so divide concentration by 10 again.

Further help:

https://socratic.org/questions/how-do-you-do-serial-dilution-calculations :smile:

Reply 7

anona1255

OP

5

Original post by aytuiq

Explanation to answer :

1cm^3 2M HCl added to 9cm^3 of water means there is a final volume of 10cm^3 with a dilution of 1:10 and 2M/10=0.2M,
This is repeated with 9cm^3 again, so divide concentration by 10 again.

Further help:

https://socratic.org/questions/how-do-you-do-serial-dilution-calculations :smile:

Thank you so much! You too :smile:

Reply 8

Namih

5

Original post by anona1255

Ah thank you, but how would you apply this to this question in particular?