1.00 *10^2 "mL" of a "12.4-M" "HCl" solution is added to water to bring the total volume of the solution to "0.820 L". What is the concentration of this new solution?

Mar 19, 2018

$\text{1.51 M}$

Explanation:

You know that you're diluting $1.00 \cdot {10}^{2} \quad \text{mL}$ of a $\text{12-4M}$ hydrochloric acid solution by adding it to enough water to get the total volume to

0.820 color(red)(cancel(color(black)("L"))) * (10^3 quad "mL")/(1color(red)(cancel(color(black)("L")))) = "820. mL"

Now, when you're diluting a solution, you must keep in mind that the ratio that exists between the volume of the diluted solution and the volume of the concentrated solution is equal to the dilution factor, $\text{DF}$.

$\text{DF" = V_"diluted"/V_"stock}$

Moreover, the dilution factor is also equal to the ratio that exists between the concentration of the concentrated solution and the concentration of the diluted solution.

$\text{DF" = c_"stock"/c_"diluted}$

In your case, the dilution factor is equal to

"DF" = (820. color(red)(cancel(color(black)("mL"))))/(1.00 * 10^2color(red)(cancel(color(black)("L")))) = color(blue)(8.20)

You can thus say that the concentration of the initial solution, i.e. the concentrated solution, was $\textcolor{b l u e}{8.20}$ times higher than the concentration of the diluted solution, and so

${c}_{\text{diluted" = c_"stock}} / \textcolor{b l u e}{8.20}$

c_"diluted" = "12.4 M"/color(blue)(8.20) = color(darkgreen)(ul(color(black)("1.51 M")))

The answer is rounded to three sig figs.

So, if you start with $1.00 \cdot {10}^{3} \quad \text{mL}$ of a $\text{12.4-M}$ hydrochloric acid solution and add it to enough water to get its total volume of $\text{0.820 L}$, you will end up with $\text{0.820 L}$ of a $\text{1.51-M}$ hydrochloric acid solution.

Don't forget that when diluting strong acids, you must always add the acid to the water and not the water to the acid!