Question #9cf86

1 Answer
Apr 13, 2017


Here's what I got.


The thing to remember about dilutions is that the decrease in the concentration of the solution must match the increase in its volume.

In other words, the ratio that exists between the volume of the diluted solution and the volume of the concentrated solution must be equal to the ratio that exists between the concentration of the concentrated solution and the concentration of the diluted solution.

#color(blue)(ul(color(black)("DF" = V_"diluted"/V_"concentrated" = c_"concentrated"/c_"diluted"))) -># the dilution factor

In your case, the concentration of the solution must decrease by a factor of

#"DF" = (6 color(red)(cancel(color(black)("M"))))/(2color(red)(cancel(color(black)("M")))) = color(blue)(3)#

which implies that the volume of the solution must increase by the same factor.

You can thus say that you have

#"DF" = V_"diluted"/V_"concentrated" implies V_"diluted" = "DF" xx V_"concentrated"#

which will get you

#V_"diluted" = color(blue)(3) xx "35 mL" = "105 mL"#

Therefore, you can say that your sample of #"6 M"# hydrochloric acid solution must be added to enough water to get the total volume of the solution to #"105 mL"#.

Remember, make sure to always add acid to water and never the other way around!