# Question 26712

Jan 23, 2017

Here's what I got.

#### Explanation:

The first thing to do here is to figure out exactly how much hydrochloric acid you get in $\text{25 mL}$ of 15% solution

25 color(red)(cancel(color(black)("mL solution"))) * "15 g HCl"/(100color(red)(cancel(color(black)("mL solution")))) = "3.75 g HCl"

Now, if you take $x$ to be the number of grams of hydrochloric acid coming from the 5% solution and $y$ to be the number of grams of hydrochloric acid coming from the 20% solution, you can say that

$x + y = 3.75 \text{ } \textcolor{\mathmr{and} a n \ge}{\left(1\right)}$

For the 5% solution, the volume that contains $x$ $\text{g}$ of hydrochloric acid is equal to

x color(red)(cancel(color(black)("g HCl"))) * "100 mL solution"/(5color(red)(cancel(color(black)("g HCl")))) = (20x)" mL solution"

For the 20% solution, the volume that contains $y$ $\text{g}$ of hydrochloric acid is

y color(red)(cancel(color(black)("g HCl"))) * "100 mL solution"/(20color(red)(cancel(color(black)("g HCl")))) = (5y)" mL solution"

You can now say that

$20 x + 5 y = 25 \text{ } \textcolor{\mathmr{and} a n \ge}{\left(2\right)}$

Use equation $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ to find the value of $x$ in terms of $y$

$x = 3.75 - y$

Plug this into equation $\textcolor{\mathmr{and} a n \ge}{\left(2\right)}$ to find the value of $y$

$20 \cdot \left(3.75 - y\right) + 5 y = 25$

$75 - 20 y + 5 y = 25$

$- 15 y = - 50 \implies y = \frac{- 50}{- 15} = \frac{10}{3}$

Use this to find the value of $x$

$x = 3.75 - \frac{10}{3} = \frac{15}{4} - \frac{10}{3} = \frac{5}{12}$

Therefore, you can say that in order to get $\text{25 mL}$ of 15%# hydrochloric acid solution, you must mix

$20 \cdot x = 20 \cdot \frac{5}{12} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{8.3 mL of 5% HCl solution}}}}$

and

$5 \cdot y = 5 \cdot \frac{10}{3} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{16.7 mL of 20% HCl solution}}}}$

I'll leave the answers rounded to two and three sig figs, respectively, but keep in mind that you only have one significant figure for the concentrations of the two stock solutions.