Question

If I have a flask that has solution containts ( 2 ml of sugar solution , 0.05 ml phenol and 5 ml sulfric acid). how I can make 1:1 dilution ? lets say I will use 250 volemtric flask

Answer 1

You haven’t mentioned the concentrations of any of the components present in the flask; hence, it isn’t possible to determine the concentrations of these components in the final solution.

A 1:1 dilution is simply a situation where you fill a volumetric flask with half analyte and half diluent. Suppose you have this 250 mL volumetric flask and say you want to prepare a 1:1 dilute solution of an analyte, say A. Suppose you have a stock solution of A (supplied by your TA or otherwise) that has a concentration of say 10 g/mL A. A 1:1 dilution will lower the final concentration of A to half the initial value, i.e, 5 g/mL. Thus, it becomes easy to find out how much of A is required and how much diluent (say water) must be added.

Use the dilution equation M₁*V₁ = M₂*V₂ where M₁ = concentration of the stock analyte = 10 g/mL; M₂ = concentration of the analyte in the prepared solution = 5 g/mL and V₂ = final volume of the dilute solution = 250 mL. Therefore,

(10 g/mL)*V₁ = (5 g/mL)*(250 mL)

====> V₁ = 125 mL.

Therefore, in order to effect a 1:1 dilution, we shall take 125 mL of the stock solution and add (250 – 125) mL = 125 mL DI water. Note that ratio of the volumes of the analyte and the diluent is 125 mL: 125 mL = 1:1 and hence, the name.