I should've listened to Ms Johnson.........

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Nowland Farms

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I should've listened to Ms Johnson who was my 5th grade Math teacher. Apparently I, either didn't listen or didn't understand so I need your input.

Here is what I need help with -

Problem - I have a liquid solution which is 7.9% active ingredient. I need to dilute this solution so that I end up with a solution which is 0.05% active ingredient.

Question - How many ounces of water do I add to 1 ounce of my solution to end up with a 0.05% mixture?

If my math is correct, I am coming up with 158 ounces of water to 1 ounce of solution. But this just doesn’t look right.

Your help is needed since I'm too embarrassed to call Ms Johnson since it has been 38 years. :oops: :oops:
 

bandit80

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I think you are right. Think algebra.

7.9 = .05x So, 7.9/.05 = x . Therefore x = 158. Meaning 1 ounce of soution to 158 ounces of dilutent (water).

I agree that doesn't sound right, should it maybe be .5 rather than .05?

This might help, but it is all in metric and you have to convert it. http://www.restrictionmapper.org/dilutioncalc9.htm

1 gallon = 3.785 liters

1 ounce = 29.57 ml
 

cfpinz

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If you really want to get picky, you actually need to add 157 ounces of water to the 1 ounce you already have to end up with a total of 158 ounces.
 

TexasBred

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Heck to keep it simple just get you about a gallon of water. Pick up your 7.9% solution and put a big squirt into the water. Stir well....should come out somewhere between .05% and .25%. Looks pretty close to me. :lol: :lol: :lol:
 

cfpinz

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fourstates":1p87evkv said:
Oh pinz, you put your foot in it! That's not correct in this equation.

You've already got .079oz of active ingredient in your one ounce sample (assuming we're using ounces as weight rather than volume). So if you set it up .079/x=.0005 and solve for x, you wind up with 158 ounces. 158 is the total number of ounces in the final product, so you must subtract the ounce you already have which supplied the .079oz of active ingredient, leaving 157oz of water to add. Not that any of that really amounts to a hill of beans, but if I'm wrong please explain why.
 

1982vett

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cfpinz":3nfwk9x5 said:
fourstates":3nfwk9x5 said:
Oh pinz, you put your foot in it! That's not correct in this equation.

You've already got .079oz of active ingredient in your one ounce sample (assuming we're using ounces as weight rather than volume). So if you set it up .079/x=.0005 and solve for x, you wind up with 158 ounces. 158 is the total number of ounces in the final product, so you must subtract the ounce you already have which supplied the .079oz of active ingredient, leaving 157oz of water to add. Not that any of that really amounts to a hill of beans, but if I'm wrong please explain why.

Maybe you end up with 159 ounces. 158 ounces water + 1 ounce solution. :?: :?
 

fourstates

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Because he is looking for a .05% solution, not a total number of ounces. You will obtain 159 oz. of the final product. If you mixed 157ozs of water you will end up with a slightly stronger solution then the .05% Nowland was looking for.
 
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Nowland Farms

Nowland Farms

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Thanks Folks, I don't think it will matter much if I add 157 or 158 ozs of water. I now remember why I didn't pay much attention to the 5th grade math. Ms Johnson was only about 20 years older than us students, she was "curvey" and wore tight sweaters. I do remember sweaters and a couple of other things very well. :dunce: :dunce: :dunce: :dunce: :dunce: :dunce:
 

Jogeephus

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Nowland Farms":ekwfy4wz said:
Thanks Folks, I don't think it will matter much if I add 157 or 158 ozs of water. I now remember why I didn't pay much attention to the 5th grade math. Ms Johnson was only about 20 years older than us students, she was "curvey" and wore tight sweaters. I do remember sweaters and a couple of other things very well. :dunce: :dunce: :dunce: :dunce: :dunce: :dunce:

Man were you lucky! My MRS. JOHNSON was a big bubble butted bee hive wearing witch who kept telling me I needed to become an astronaut because all I did was take up space in her world. (She was a smart lady) :oops:
 

cfpinz

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fourstates":2unrvcpq said:
Because he is looking for a .05% solution, not a total number of ounces. You will obtain 159 oz. of the final product. If you mixed 157ozs of water you will end up with a slightly stronger solution then the .05% Nowland was looking for.

Not that the one ounce will amount to a darn thing in the big picture, but considering the initial 1oz contains .079oz of product, look at it this way:

.079oz divided by 158 total ounces equals .05% concentration.

.079oz divided by 159 total ounces equals .0496855...% concentration.

Arguements?
 

peg4x4

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What are we diluteing? Is it poisonious if not done exactly right? What happens if the diluteing is wrong??
 

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