Question

Could you please check my work and let me know if my answer is correct? I don't have a solutions guide and I want to make sure I'm really comprehending the material.

<b>Calculate the # of grams of barium flouride that will dissolve in 2.50 L of pure water at 25 degrees C. Assume the Ksp of barium flouride us given as 1.7 x 10^-6.</b>

here is my work:

BaF2 (arrows in both directions) Ba (aq) + 2 F (aq)

ksp = [Ba][F]^2

let x= [Ba]
2x= [F]

1.7x10^-6 = (2x)^2(x)
x = ((1.7x10^-6)/4)
x= 1.4 x 10^-7

1.4 x 10^-7 mol Ba/ L (2.50 L)(1 mol BaF2/ 1 mol Ba)(175.33g/1 mol BaF2) = 438.32 g
or 438 g

thanks in advance!

Answer 1

Calculate the # of grams of barium flouride that will dissolve in 2.50 L of pure water at 25 degrees C. Assume the Ksp of barium flouride us given as 1.7 x 10^-6.

here is my work:

<b>Thanks for showing your work. See comments below.</b>

BaF2 (arrows in both directions) Ba (aq) + 2 F (aq)

ksp = [Ba][F]^2

let x= [Ba]
2x= [F]

1.7x10^-6 = (2x)^2(x) <b>OK to here.</b>
x = ((1.7x10^-6)/4) <b>
x³ = (1.7E-6/4)
x = [(1.7E-6)/4]^1/3 = 7.52E-3 Mols/L. Convert this to 2.5L and to grams.</b>
x= 1.4 x 10^-7

1.4 x 10^-7 mol Ba/ L (2.50 L)(1 mol BaF2/ 1 mol Ba)(175.33g/1 mol BaF2) = 438.32 g
or 438 g

thanks in advance!

Answer 2

thank you! i didn't realize i never took the cube root.