What is the concentration of the solution produced when 150.0 mL of water is added to 200.0 mL of 0.250 M NaCl?

1 Answer
Feb 13, 2018

Answer:

#"0.143 M"#

Explanation:

The thing to remember about a dilution is that you can find its dilution factor by dividing either the volume of the diluted solution by the volume of the stock solution

#"DF" = V_"diluted"/V_"stock"#

or by dividing the concentration of the stock solution by the concentration of the diluted solution.

#"DF" = c_"stock"/c_"diluted"#

So you can say that for any dilution, you have

#"DF" = V_"diluted"/V_"stock" = c_"stock"/c_"diluted"#

In your case, the volume of the diluted solution will be

#V_"diluted" = "150.0 mL + 200.0 mL" = "350.0 mL"#

This means that the dilution factor is equal to

#"DF" = (350.0 color(red)(cancel(color(black)("mL"))))/(200.0 color(red)(cancel(color(black)("mL")))) = color(blue)(1.75)#

You can thus say that the stock solution was #color(blue)(1.75)# times more concentrated than the diluted solution, since

#c_"stock" = "DF" * c_"diluted"#

This implies that the concentration of the diluted solution is equal to

#c_"diluted" = c_"stock"/"DF"#

#c_"stock" = "0.250 M"/color(blue)(1.75) = color(darkgreen)(ul(color(black)("0.143 M")))#

The answer is rounded to three sig figs.