Calculating Amount of Pure Acid Needed for Desired Solution

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The discussion focuses on calculating the amount of pure acid needed to create a 30% acid solution from a 10% acid solution. The chemist needs to mix 5 milliliters of a 10% acid solution with an unknown volume of pure acid (100% concentration). The correct equation derived is 0.10(5) + x = 0.30(x + 5), where x represents the volume of pure acid in milliliters. This equation effectively models the problem and allows for the calculation of the required amount of pure acid.

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mathdad
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A chemist needs a 30% acid solution. How much pure acid needs to be mixed with 5 milliliter a of a 10% acid solution to obtain the desired solution?

The words PURE ACID for some reason or another point to 100 percent.

My equation:

0.30(x) + 0.10(5) = (x + 5)

Is this right?
 
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What you want, where x is the amount of pure acid added in mL:

0.10(5) + x = 0.50(x + 5)
 
MarkFL said:
What you want, where x is the amount of pure acid added in mL:

0.10(5) + x = 0.50(x + 5)

One of my greatest struggles in math is to create an equation(s) from the words in applications. It is clearly the most important, in my opinion, math skill to learn. I vividly recall getting rejected for a Financial Advisor job in 2006 because I simply could not pass the bank test, which consisted mostly of word problems. The F. A. job could have changed my finances for the better. I took this test (and other exams) after earning my B.A. degree in 1994.
 

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