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#### scarmcadds

##### New member

- Aug 13, 2020

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- Thread starter scarmcadds
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- Aug 13, 2020

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- Mar 1, 2012

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initial mixture is $\dfrac{16}{35} \approx 45.7$% acid and $\dfrac{19}{35} \approx 54.3$% water

add x units of acid ...

(16+x) units of acid + 19 units water = (35+x) units of new mixture

need to add x amount of acid such that $\dfrac{16+x}{35+x} = \dfrac{19}{35}$ and $\dfrac{19}{35+x} = \dfrac{16}{35}$

solving either equation for x will give the requisite number of original units of acid to be added ... just recall the question asks for the added x as percentage of the original number of units in the starting mixture.