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Homework Help: Confusing Algebra problem help

  1. Mar 18, 2005 #1
    Ok, I have this word problem dealing with exponential growth and decay. I need serious help. There are three parts to this problem...I really need help with this.

    Here's the problem:

    A dilution is commonly used to obtain the desired concentration of a sample. For example, suppose that a 1 millilitre of hydrochloric acid, or HCl, is combined with 9 millilitres of a buffer. The concentration of the resulting mixture is 1/10 of the original concentration of HCl.

    a.) Suppose that this dilution is performed again with 1 millilitre of the already diluted mixture and a 9 millilitre buffer. What is the concentration of the resulting mixture (compared with the original concentration)?

    b.) Write an expression to model the concentration of HCl in the resulting mixture after repeating dilusions as described in part "a".

    c.) What is the concentration of the resulting mixture (compared to the orginal concentration) after 5 repeated dilusions?

    Ok, I don't even know what I'm supposed to find or how I'm supposed to find it. I tried this in class without any luck, unfortunately my calculations only took me as far as .1^.1. I don't even think thats right. Could someone please help me? I really need it. Thanks.
  2. jcsd
  3. Mar 18, 2005 #2
    well I can't figure an answer right now, but it looks as though you will need to use [tex] y=mx^2+b[/tex]
  4. Mar 18, 2005 #3
    Just out of curiosity, why would I use y=mx+b? I was following the amount x multiplier^to a power. I don't know what to do.
  5. Mar 18, 2005 #4
    Well you can calculate the initial HCl concentration : molar mass HCl/volume in litre

    Now after each dilusion, the initial concentration is divided by ten right?

    Sow you would get something like : inital concentration is denoted by C0

    concentration C = (C0)*(10)^(-n) where n denotes the number of dilusions

  6. Mar 18, 2005 #5
    Ummm, right. :confused:

    What I posted is what I've got for information. Would I divide the molar mass by 1000? I would assume each dilusion would be the intial concentration divided by 10. I don't understand the CO notation you're talking about. Wow, I suck at word problems! Thanks Marlon.

    I mean my thank you seriously, it isn't ment to be sarcastic or deterrogatory. I really do suck at word problems. :frown:
    Last edited: Mar 18, 2005
  7. Mar 18, 2005 #6


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    You got your first concentration by multiplying a pure concentration (1) by .1 {1 ml divided by (1ml + 9ml)}.

    You dilute the resulting concentration by the same amount, or you're multiplying .1 by .1.

    In other words, you have concentration (C) of:

    [tex]C = 1 * .1 * .1 * .1 ..... [/tex] (You multiply by .1 each time you dilute it.

    If you want to simplify it into a formula, you're multiplying 1 by [tex].1^n[/tex] where n states how many times you're performing the dilution. It's not .1, since you can't perform the dilution .1 times (unless you hear the phone ring as soon as you start to poor in the buffer, I guess).

    I think accidently sniffing their chemicals must be an occupational hazard of being a chemist. Chemists are often 'dilusional'. :rofl:
  8. Mar 18, 2005 #7
    Co just denotes the initial concentration that you calculate. the molar mass of elements is given in the tabel of Mendelejev

  9. Mar 18, 2005 #8
    got it thanks
  10. Mar 19, 2005 #9
    i just put in [tex]y=mx^2+b[/tex] just in case you had a calculator, that could then make the graph for you and determine the exponential decay, i spose the more proper way to show it would be [tex] y=mx^n [/tex]
  11. Mar 19, 2005 #10
    Ah ok, I think I'm more graphing calculator illiterate than I am computer illiterate. :redface:
  12. Mar 21, 2005 #11


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    [itex] x^{n} [/itex] is not exponential decay!!.
  13. Mar 21, 2005 #12
    Notice for each dilution your initial dilution D is decreasing by a power of 10

    Original concentration = 1ml (100%)
    First dilution -> 0.1ml (10%)
    Second dilution -> 0.01ml (1%)
    nth Diltuion -> D x 10^(-n)

    Concentration C is proportional to original dilution D and inversely proportional to 10 raised to the number of dilutions

    C = D/(10^n)
    So for 5 dilutions you would have C = D/10^5, where D = 1ml

    C = 0.00001ml or 1 x 10^-5 ml
    Last edited: Mar 21, 2005
  14. Mar 21, 2005 #13
    Whozum, thanks! I just worked out your formula with the data and checked the answer in the back of the book, its correct. Thanks for the help everyone! :biggrin:
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