00:02
Here in this problem, we have to find out the mass of barium sulfate that can dissolve in 5 liter of water.
00:10
Now, let's write the relevant equilibrium equation for barium sulfate.
00:16
B .a.
00:17
So4, solid barium sulphate is in equilibrium with its ion, barium 2 plus in solution, aqua solution, plus sulfate ion, so4, 2, much.
00:33
Minus aqua solution.
00:36
Now, uh, the, let's write the solubility product expression.
00:42
Ksp is concentration of barium two plus times concentration of so4 2 minus.
00:52
Now, here one more barium sulfate producing one mole ba2 plus and one more so4 2 minus.
01:00
Let the molar solubility for, barium sulphate b s so molar solubility for this ion also s and for sulphate iron also will be now the kasp we get from the chart from the table for barium sulphate is 1 .1 times 10 to the power negative 10 that is equal to s times s so s square is 1 .1 times 10 to 1 .1 times 10 to the power negative 10 so s will be square root of 1 .1 times 10 to the power negative 10 so the value for s will be 1 .05 times 10 to the power negative 5 molar so the molar solubility of barium sulfate is therefore molar solubility molar solubility of barium sulphate will be 1 .05 times 10 to the power minus 5 more per liter that is that much molar...