In order to make 1200mL of 2.0 M of sodium hydroxide, NaOH, from 9M NaOH, how many mL of the 9M NaOH should be used?
mL1 x M1 = mL2 x M2
Wow....how did I NOT see that...thanks DrBob... :-)
To solve this problem, we can use the formula for dilution:
(C1)(V1) = (C2)(V2)
Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
In this case, the initial concentration (C1) is 9 M, the final concentration (C2) is 2 M, and the final volume (V2) is 1200 mL.
We need to find the initial volume (V1) of the 9 M NaOH that should be used.
Substituting the values into the formula, we have:
(9 M)(V1) = (2 M)(1200 mL)
To solve for V1, divide both sides of the equation by 9 M:
V1 = (2 M)(1200 mL) / 9 M
Now, we can calculate V1:
V1 = 2400 mL / 9
V1 ≈ 266.67 mL
Therefore, approximately 266.67 mL of the 9 M NaOH should be used to make 1200 mL of 2 M NaOH.