Please show all steps, Thanks!!! 0.7500 L of a solution that contained 480.4 ppm

Question

Please show all steps, Thanks!!!

0.7500 L of a solution that contained 480.4 ppm Ba(NO3)2 was mixed with 0.2000 L of a 0.03090 M solution of Al2(SO4)3 . BaSO4 precipitates and settles out of solution. Using this information,

Find the concentrations of each of the ions left in solution after precipitation has finished: Ba ], [S04 ], [Al], INO3]. Assume all volumes are additive and again neglect equilibrium issues, such as Kop Attach extra paper as needed issues, such as Ksp- Attach extra paper as needed.

Explanation / Answer

First we have to figure out how many moles of each reactant we have and then look at the balanced equation to see how many moles of BaSO4 would form.

We know that "ppm" is the same thing as "mg/L."

480.4 mg Ba(NO3)2 / L x 0.7500 L = 360.3 mg Ba(NO3)2

360.3 mg Ba(NO3)2 x (1 g / 1000 mg) x (1 mole Ba(NO3)2 / 261.4 g Ba(NO3)2) x (1 mole Ba / 1 mole Ba(NO3)2) = 0.001378 moles Ba

similarly

360.3 mg Ba(NO3)2 x (1 g / 1000 mg) x (1 mole Ba(NO3)2 / 261.4 g Ba(NO3)2) x (2 mole NO3 / 1 mole Ba(NO3)2) = 0.001378 moles Ba

= 0.002756 moles NO3

moles Al2(SO4)3 = M Al2(SO4)3 x L Al2(SO4)3 = (0.03090)(0.2000) = 0.006180 moles Al2(SO4)3

0.006180 moles Al2(SO4)3 x (3 moles SO4 / 1 mole Al2(SO4)3) = 0.01854 moles SO4

Similarly

0.006180 moles Al2(SO4)3 x (2 moles Al / 1 mole Al2(SO4)3) = 0.01236 moles Al3+

The balance equation for above reaction is 3Ba(NO3)2 + Al2(SO4)3 ---> 2Al(NO3)3 + 3BaSO4

The net equation is Ba2+(aq) + SO4 2-(aq) ==> BaSO4(s)

Hence, Ba and SO4 react in a 1:1 mole ratio. We have available 0.001378 moles Ba and 0.01854 moles of SO4. Obviously. The reaction can be summarized:

Moles . . . . . .Ba2+ . .+ . .SO4 2- . .==> . .BaSO4
Initial . . . . .0.001378 . . .0.01854 . . . . . . . . .0
Change . . -0.001378 . .-0.001378 . . . . . . +0.001378
Final . . . . . . . 0 . . . . . . 0.01716 . . . . . . .0.001378

So we formed 0.001378 moles of BaSO4.

Now let us calculate remaining ions in solution:

1) SO42-

    The final volume is 750.0 mL + 200.0 mL = 950.0 mL = 0.9500 L

    0.01716 moles of SO42- / 0.95 L = 0.01806 M

2) Al3+

0.01236 moles/0.95 L = 0.01301 M

3) NO3-

    0.002756 moles/0.95 L = 0.0029 M

4) Ba2+ = 0 M

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