Calculate dilution chemistry?
Hello, I have a question about the following task:
When preparing the reagents for determining water hardness, 3 liters of 4.5 M sodium hydroxide solution were mistakenly prepared. How much purified water should be used to dilute the existing sodium hydroxide solution to achieve the required concentration of 2 mol/l?
My approach: You calculate the amount of NaOH (3L * 4.5 mol/L = 13.5 mol). Then you need the volume of distilled water, which I would have calculated as follows: 2 mol/L = 13.5 mol / V. This gives 6.75 L, but the solution says the result is 3.75 liters. Where did I go wrong?
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3 L with the concentration c=4.5 mol/L contain a total amount of n=13.5 mol. A concentration of c=2 mol/L is required.
c = n/V ==> V = n/c
V = 13.5 mol/(2 mol/L) = 6.75 L
Therefore, the 3 L must be with demin. water to 6.75 L. This is somewhat different than 3.75 L of water added, since in such relatively concentrated solutions the volumes do not necessarily be additive (volume contraction).