#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
Explanation:
You know that you're diluting
#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" = 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
#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
Don't forget that when diluting strong acids, you must always add the acid to the water and not the water to the acid!