Question

A standardized solution that is 0.0500 0.0500 M in Na + Na+ is necessary for a flame photometric determination of the element. How many grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 800.0 mL of this solution?

Answer 1

2.12 grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 mL of this solution.

Step-by-step explanation:

Molarity of sodium ions =
`[Na^+]` = 0.0500 M

Moles of sodium ions = n

Volume of the solution = V = 800.0 mL = 0.800 L

`Molarity=(n)/(V(L))`

`[Na^+]=(n)/(V)`

`n=[Na^+]* V=0.0500 M* 0.800 L=0.04 mol`

`Na_2SO_4(aq)\rightarrow 2Na^+(aq)+CO_3^(2-)(aq)`

1 mole sodium carbonates gives 2 moles of sodium ion and 1 mole of carbonate ions.

Then 0.04 moles of sodium ions will be obtained from:

`(1)/(2)* 0.04 mol=0.02 mol` of sodium m carbonation.

Mass of 0.02 moles of sodium carbonate = 0.02 mol × 106 g/mol= 2.12 g

2.12 grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 mL of this solution.