"Na"_3"PO"_4Na3PO4 dissolves in water to produce an electrolyte solution. What is the Osmolarity of a 2.0 * 10^(-3)"M Na"_3"PO"_42.0⋅10−3M Na3PO4 solution? Thank you!
1 Answer
Explanation:
The thing to keep in mind about osmolarity is that it takes into account the number of moles of particles of solute that are produced in a solution when a given number of moles of solute are dissolved to make said solution.
In other words, you can think about osmolarity as being a multiple of molarity
color(blue)(ul(color(black)("osmolarity" = i xx "molarity")))
Here
In your case, trisodium phosphate is a strong electrolyte, which implies that it dissociates completely in aqueous solution to produce sodium cations and phosphate anions
"Na"_ 3"PO"_ (4(aq)) -> 3"Na"_ ((aq))^(+) + "PO"_( 4(aq))^(-)
Now, notice that every mole of trisodium phosphate that dissociates in solution produces a total of
"3 moles Na"^(+) + "1 mole PO"_4^(-) = "4 moles ions"
The number of moles of particles of solute produced in solution are actually called osmoles.
As a result, the van't Hoff factor will be equal to
i = "4 moles ions produced (osmoles)"/("1 mole Na"_3"PO"_4color(white)(.)"dissolved") = 4
Since you know that
["Na"_3"PO"_4] = 2.0 * 10^(-3)"M"
you can say that the solution will have an osmolarity equal to
color(darkgreen)(ul(color(black)("osmolarity" = 4 xx 2.0 * 10^(-3)"M" = 8.0 * 10^(-3)color(white)(.)"osmol L"^(-1))))
It's important to keep in mind that osmolarity is expressed in osmoles per liter because you have
(2.0 * 10^(-3)color(red)(cancel(color(black)("moles Na"_3"PO"_4))))/"1 L solution" * "4 osmoles"/(1color(red)(cancel(color(black)("mole Na"_3"PO"_4)))) = 8.0 * 10^(-3)color(white)(.)"osmol L"^(-1)
