If somehow, #"50.0 mL"# of a stock solution of #"120. M"# #"HCl"# is available, what volume of water should the appropriate amount of #"HCl"# be added to, in order to reduce its concentration to #"4.00 M"#?
1 Answer
I got
Just so you know, that initial concentration of
I assume you mean the "before/after" formula:
#\mathbf(M_1V_1 = M_2V_2)#
Basically, you have a relationship of concentration to volume:
- If concentration is to increase, then volume has to decrease.
- So, if volume increases, then concentration must decrease (the contrapositive).
Here,
For the volume, we can use milliliters (
What you already have are:
#V_1 = "50.0 mL"# #M_1 = "120. M"# #M_2 = "4.00 M"#
So, you are solving for the final volume,
#color(green)(V_2) = ((M_1)/(M_2)) V_1#
#= (("120." cancel"M")/(4.00 cancel"M")) ("50.0 mL")#
#=# #"1500 mL"#
Or, to three sig figs:
#= color(green)(1.50xx10^3)# #color(green)("mL")#
Remember, that's your FINAL volume, NOT the amount of water you might start with, so you should subtract to get:
It's better, however, to start with the
You won't necessarily transfer exactly