Solutions

[Aya Watson 2B]

Hi! I'm stuck on G.5. There was a post about it earlier, but I couldn't understand the explanation. Would someone be able to write out their work please?

A student prepared a solution of sodium carbonate by adding 2.111 g of the solid to a 250.0-mL volumetric flask and adding water to the mark. Some of this solution was transferred to a buret. What volume of solution should the student transfer into a flask to obtain (a) 2.15 mmol Na+ (b) 4.98 mmol CO3^2- (c) 50.0 mg Na2CO3?

[Brian Diehl 2B]

Hey Aya!

Before tackling parts a-c, I first calculated the molarity of the Na₂CO₃ solution being transferred to the buret:

(2.111 g Na₂CO₃)(1 mol/105.99 g) = 0.01992 mol Na₂CO₃

M = mol/L = (0.01992 mol/0.2500 L) = 0.07968 mol/L Na₂CO₃

The fact that molarity is a ratio between moles and liters will be useful when performing the stoichiometry needed to find the volumes of transferred solution for parts a-c. Always start with your given value (mmol or mg), convert to moles of Na₂CO₃, and then find the volume of Na₂CO₃ using the molarity ratio (0.07968 mol Na₂CO₃/1 L Na₂CO₃).

a)

(2.15 mmol Na⁺)(1 mol/1000 mmol)(1 mol Na₂CO₃/2 mol Na⁺)(1 L Na₂CO₃/0.07968 mol Na₂CO₃) = 0.0135 L = 13.5 mL Na₂CO₃

b)

(4.98 mmol CO₃^2-)(1 mol/1000 mmol)(1 mol Na₂CO₃/1 mol CO₃^2-)(1 L Na₂CO₃/0.07968 mol Na₂CO₃) = 0.0625 L = 62.5 mL Na₂CO₃

c)

(50.0 mg Na₂CO₃)(1 g/1000 mg)(1 mol/105.99 g)(1 L Na₂CO₃/0.07968 mol Na₂CO₃) = 0.00592 L = 5.92 mL Na₂CO₃

Hope this helps!