#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?

1 Answer
Mar 19, 2018

Answer:

#"1.51 M"#

Explanation:

You know that you're diluting #1.00 * 10^2 quad "mL"# of a #"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, #"DF"#.

#"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.

#"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 #color(blue)(8.20)# times higher than the concentration of the diluted solution, and so

#c_"diluted" = c_"stock"/color(blue)(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 * 10^3 quad "mL"# of a #"12.4-M"# hydrochloric acid solution and add it to enough water to get its total volume of #"0.820 L"#, you will end up with #"0.820 L"# of a #"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!