how many grams of MgSO4 can be added to the solution before BaSO4 begins to precipitate ?

Question

Starting with 14.8 ml of 3.30 × 10−2 M BaI2. So again, how many grams of MgSO4 can be added to the solution before BaSO4 begins to precipitate ?

Answer 1

►BaI2 --> Ba+2 + 2I - (100% dissociate)
►MgSO4 --> Mg+2 + SO4-2 (100% dissociate)

Ba+2 will react with SO4-2 in the formation of BaSO4;
for finding how many grams of MgSO4 can be added to the solution before BaSO4 begins to precipitate; we need solubility product (Ksp) of BaSO4 which is 1.5x 10 ^-9,
so;
► Ksp = 1.5x 10 ^-9 = [Ba+2][SO4-2]

we know that the conc. of Ba+2 is equal to the conc. of BaI2 which is 3.3x10^-2
so
1.5x 10 ^-9 =3.3x10^-2 *[SO4-2]
=> [SO4-2] =0.45x 10 ^-7

this conc is equal to the conc. of MgSO4 and also we know that the volume change due to addition of MgSO4 is negligible so;

no of moles of MgSO4 = 0.45x 10 ^-7 * 14.8 x 10 ^-3
no of moles of MgSO4 = 6.66 x10^-10

multiplying no of moles in molucular weight of no of moles of MgSO4 will give us the grams of no of moles of MgSO4 :

120 *6.66 x10^-10 = 799.2 x10^-10 gr

if more than this amount is added BaSO4 will precipitate.

hope this helps

Answer 2

I had the exact same question as you did (except for the numbers) and the Ksp value that worked for me was 1.1x10^-10. So the answer is this number divided by 0.033 and then multiplied by (14.8/ 10000) and then multiplied by the molar mass, 120.3676. ans= 5.938e-9 (i think) hope that helps

Answer 3

that process is right but the Ksp is actually 1.1e-10