SOLUTION: What amount of pure acid must be added to 500 mL of a 25% acid solution to produce a 50% acid solution?

Question 1200052: What amount of pure acid must be added to 500 mL of a 25% acid solution to produce a 50% acid solution?

Found 3 solutions by ikleyn, greenestamps, josgarithmetic:
Answer by ikleyn(52786)   (Show Source): You can put this solution on YOUR website!
.
What amount of pure acid must be added to 500 mL of a 25% acid solution
to produce a 50% acid solution?
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Let V be the volume of the pure acid to add, in milliliters. Then the total volume of the mixture is (500+V) mL, and the volume of the pure acid is (0.25*500+V). Our concentration equation is = 0.50 (which is 50%). It gives 0.25*500 + V = 0.5*(500+V) 0.25*500 + V = 0.5*500 + 0.5V V - 0.5V = 0.5*500 - 0.25*500 0.5V = 125 V = = 250. ANSWER. 250 mL of the pure acid should be added.

Solved.

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It is a standard and typical mixture problem.

For introductory lessons covering various types of mixture word problems see
    - Mixture problems
    - More Mixture problems
    - Solving typical word problems on mixtures for solutions
    - Typical word problems on mixtures from the archive
in this site.

You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization, from very detailed to very short.

Read them and become an expert in solution mixture word problems.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook in the section "Word problems" under the topic "Mixture problems".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

The response from the other tutor shows a standard formal algebraic solution -- which you should understand, if you are a student learning how to solve problems using algebra.

If the speed of obtaining the answer is important and any method can be used (as in a timed math competition), here is an informal solution method that is much faster and easier than the formal algebraic method.

The 50% target concentration is 1/3 of the way from the 25% concentration of the original acid solution to the 100% concentration of the acid that is added. (25 to 50 is a difference of 25; 25 to 100 is a difference of 75; 25/75 = 1/3.)

That means 1/3 of the mixture should be what is being added.

So 2/3 of the final mixture is the original 25% acid solution; and that means the amount of 100% acid being added is half the amount of the original 25% acid solution.

ANSWER: 1/2 of 500 mL = 250 mL

Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!

MATERIALS CONC.% VOLUME PURE Pure Acid 100 v 100v Lower Conc. 25 500 (25)(500) ResultingMix 50 v+500 (50)(v+500)

The amount of pure acid in the parts to be mixed is the same as the pure acid of the resulting mixing.

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----------simplify and compute.
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