Question

What is the concentration in molarity of S2O32- (aq) in a solution prepared by mixing 300 mL of 0.162 M Na2S2O3 (aq) with enough 0.126 M KCl (aq) to make a 350.0 mL solution? (Assume volumes are additive and that no chemical reactions take place.)

Answer 1

`[Na_2C_2O_3]_"final"` = `0.139M`

Explanation:

If no reaction takes place (which is true in the case noted), you're simply diluting the concentrate to (30/35)th of it's original concentration to give a `(30)/(35)xx0.162M` = 0.139M Solution of `Na_2S_2O_3`.

In solution, the thiosulfate ion concentration will be equal to the final concentration of the diluted sodium thiosulfate because it's a 1 to 1 ionization ratio. That is,

`Na_2S_2O_3` => `2Na^+(aq)` + `S_2O_3^(2-)(aq)`
`("1 mole")` `color(white)(xxxx)("2 mole")` `color(white)(xxx)("1 mole")`

For determining concentration of diluted solution, one may also use 'The Dilution Equation' => Molarity x Volume of Concentrated Solution = Molarity x Volume of Diluted Solution, or ...

`(Molarity xx Volume)_"concentrate"`

= `(Molarity xx Volume)_ "diluted"`

`(0.162M)(300ml)` = `[ S_2O_3^(2-)]"(350ml)"`

=> `[S_2O_3^(2-)]` = `(0.162M)(300cancel(ml))/(350cancel(ml))` = `0.139M`