Question

Titrations with hydrated solutions Can anyone do it?

Sodium carbonate exists in hydrated form, `"Na"_2"CO"_3cdotx"H"_2"O"`, in the solid state. `"3.5 g"` of a sodium carbonate sample was dissolved in water and the volume made up to `"250 cm"^3`. `"25.0 cm"^3` of this solution was titrated against `"0.1 mol/dm"^3` `"HCl"` and `"24.5 cm"^3` of the acid were required. Calculate the value of x given the equation:

`Na_2CO_3(aq) + 2HCl(aq) -> 2NaCl(aq) + CO_2(g) + H2O(l)`

Answer 1

`x ~~ 10`

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The titration involved:

`"Na"_2"CO"_3(aq) + 2"HCl"(aq) -> 2"NaCl"(aq) + "CO"_2(g) + "H"_2"O(l)`

We know that `"24.5 cm"^3` (or `"mL"`!) of the acid was used to titrate, and that there are `"2 mols"` of `"HCl"` theoretically needed per `"1 mol Na"_2"CO"_3`.

Therefore,

`24.5 cancel"mL" xx cancel"1 L"/(1000 cancel"mL") xx "0.1 mols HCl"/cancel"L soln"`

`=` `"0.00245 mols HCl"`

were used to titrate

`0.00245 cancel"mols HCl" xx ("1 mol Na"_2"CO"_3)/(2 cancel"mols HCl")`

`= "0.001225 mols Na"_2"CO"_3`

This is a crucial step most students miss:

This was `0.001225` `mols` in `25.0` `mL`, so `0.01225` `mols` are in `250` `cm^3`. If you forget to do this, you would get `x ~~ 1`...

Knowing the mass of hydrated solid used, we can then find the mass of water in the hydrated solid.

`0.01225 cancel("mols Na"_2"CO"_3) xx ("105.986 g Na"_2"CO"_3)/cancel("1 mol Na"_2"CO"_3)`

`=` `"1.298 g anhydrous solid"`

Hence, `"3.5 g" - "1.298 g" = "2.20 g"` `x "H"_2"O"`.

Since this is for `bbx` equivalents of water, we'll need the mass of `bb1` equivalent of water on the scale of `"0.01225 mols hydrate"`.

For every `"0.01225 mols hydrate"`, `"0.01225 mols"` of water would correspond to

`0.01225 cancel("mols H"_2"O") xx "18.015 g"/cancel("1 mol H"_2"O") = "0.2207 g H"_2"O"`

That means

`("2.20 g"color(white)(.)x "H"_2"O")/("0.2207 g 1H"_2"O") = 9.97 = x`

And so, this is roughly a decahydrate.

`-> color(blue)("Na"_2"CO"_3cdot10"H"_2"O")`