How much pure acid must be added to 6 milliliters of a 5% acid solution to produce a 40% acid solution?

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How much pure acid must be added to 6 milliliters of a 5% acid solution to produce a 40% acid solution?

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Answer 1 · 2015-08-06T00:10:25

You need to add 3.5 mL of pure acid.

Explanation:

In order to solve this problem, you need to use the fact that percent concentration by volume is defined as the ratio between the volume of the solute, in your case the acid, and the volume of the solution, acid + water, multiplied by 100.

$%v/v = \frac{m_{acid}}{m_{sol}} \cdot 100$ $"%v/v" = m_"acid"/m_"sol" * 100$

Use this equation to determine how much acid your original 6-mL sample contains

$m_{acid} = \frac{v/v% \cdot m_{sol}}{100} = \frac{5 \cdot 6 mL}{100} = 0.3 mL$ $m_"acid" = ("v/v%" * m_"sol")/100 = (5 * "6 mL")/100 = "0.3 mL"$

Since you're adding pure acid to this solution, the volume of the solute and the volume of the solution will increase by the same amount, $x$ $x$ mL, so that you can write, for the target solution

$\frac{m_{acid} + x}{m_{sol} + x} \cdot 100 = 40 %$ $(m_"acid" + x)/(m_"sol" + x) * 100 = 40%$

This is equivalent to

$\frac{0.3 + x}{6 + x} \cdot 100 = 40$ $(0.3 + x)/(6 + x) * 100 = 40$

Solve this equation for $x$ $x$ to get

$(0.3 + x) \cdot 100 = 40 \cdot (6 + x)$ $(0.3 + x) * 100 = 40 * (6 + x)$

$30 + 100 x = 240 + 40 x$ $30 + 100x = 240 + 40x$

$60 x = 210 \Rightarrow x = \frac{210}{60} = 3.5 mL$ $60x = 210 => x = 210/60 = color(green)("3.5 mL")$

You thus need to add 3.5 mL of pure acid to your original solution to get its percent concetration by volume to go from 5 to 40%.

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How much pure acid must be added to 6 milliliters of a 5% acid solution to produce a 40% acid solution?

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