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.
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:
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!
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:
Thanks for showing your work. See comments below.
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) OK to here.
x = ((1.7x10^-6)/4)
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.
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!
thank you! i didn't realize i never took the cube root.
To check your work, we can compare it to the expected result and see if it matches. However, before we do that, let's go through the steps of the calculation to ensure accuracy.
1. Start by writing the balanced equation for the dissolution of barium fluoride: BaF2 (s) ⇌ Ba2+ (aq) + 2F- (aq)
2. The equilibrium expression relates the concentrations of the dissolved ions to the solubility product constant (Ksp): Ksp = [Ba2+][F-]^2
3. Let's assume the concentration of Ba2+ is x M and the concentration of F- is 2x M. Substituting these values into the Ksp expression gives: Ksp = (x)(2x)^2 = 4x^3
4. Now we can substitute the given value of Ksp (1.7 x 10^-6) into the equation: 1.7 x 10^-6 = 4x^3
5. Solve for x by rearranging the equation: x^3 = (1.7 x 10^-6) / 4
6. Calculate x using a calculator: x ≈ 4.38 x 10^-4
7. Now that we have the concentration of Ba2+ (x ≈ 4.38 x 10^-4 M), we can proceed to calculate the grams of barium fluoride that will dissolve.
8. The molar mass of BaF2 is 175.33 g/mol. Multiply the concentration of Ba2+ (x ≈ 4.38 x 10^-4 M) by the molar mass of BaF2: (4.38 x 10^-4 mol/L) * (175.33 g/mol) = 0.0767 g/L
9. Finally, multiply this value by the volume of water (2.50 L) to obtain the total grams of barium fluoride that will dissolve: (0.0767 g/L) * (2.50 L) = 0.192 g
According to our calculations, the correct answer is approximately 0.192 grams of barium fluoride that will dissolve in 2.50 L of pure water at 25 degrees C. Therefore, it seems like your answer of 438 grams is incorrect. Please double-check your calculations and make sure you followed each step correctly.