Question

What is the concentration of `"sulfate"` in a solution made from `20.7*g` `"ammonium sulfate"` in a volume of `125*mL`?

Answer 1

Approx....`1.25*mol*L^-1`....

Explanation:

Our defining equation is....

`"Moles of solute"/"Volume of solution"`.

And so with respect to ammonium sulfate, `(NH_4)_2SO_4` we gots...

`((20.7*g)/(132.14*g*mol^-1))/(125*mLxx10^-3*L*mL^-1)=1.25*mol*L^-1`

Now this is respect to ammonium sulfate....of course we know that ammonium sulfate speciates in solution to give...

`(NH_4)_2SO_4(s) stackrel(H_2O)rarr2NH_4^+ + SO_4^(2-)`

And CLEARLY we got a solution that is `1.25*mol*L^-1` with respect to sulfate ion. You should be able to tell us the concentration with respect to ammonium ion.... Why should it be double?

Of course, ammonium will undergo some water hydrolysis to give ammonia, but this should be minimal....

`NH_4^+ +H_2O(l)rightleftharpoonsNH_3(aq) + H_3O^+`

What you predict as to the `pH` of the resultant solution? Below `7`, equal to `7`, above `7`?