Equal volumes of 1 M BaCl2 and 1 M NaSO4 are mixed. What effect will the addition of more Na2SO4 solution have on the concentration of Ba^2+ remaining in solution?

Question

Equal volumes of 1 M BaCl2 and 1 M NaSO4 are mixed. What effect will the addition of more Na2SO4 solution have on the concentration of Ba^2+ remaining in solution?

*No effect
*increase the concentration of Ba^2+ ions
*Decreases the concentration of Ba^2+
*Ba^2+ will remain the same concnetration but be ionized less.
*There are no Ba^2+ ions in solution, so the Na2SO4 has nothing with which to react.

Answer 1

Yes, there is some (not much but some) Ba^+2 in solution.

BaSO4 ==> Ba^+2 + SO4^=
Ksp is about 1 x 10^-10
so (y)(y) = 1 x 10^-10
y = sqrt (1x10^-10) = 1 x 10^-5 M

So you look at the addition of Na2SO4. That is adding a common ion (remember the common ion effect) and that will shift the equilibrium to the ????? (left or right) making BaSO4 ?????(more or less) soluble.

Answer 2

BaSO4 less soluble so answer D

Answer 3

Right, BaSO4 will be less soluble but it appears that is answer C, not D. You don't have the answers labeled.

Answer 4

To determine the effect of adding more Na2SO4 solution on the concentration of Ba^2+ remaining in solution, we need to consider the reactions that occur between the ions present.

When 1 M BaCl2 and 1 M Na2SO4 are mixed, the following reaction occurs:

BaCl2 (aq) + Na2SO4 (aq) -> BaSO4 (s) + 2 NaCl (aq)

This reaction forms an insoluble white precipitate of BaSO4 and soluble sodium chloride (NaCl). The Ba^2+ ions combine with SO4^2- ions to form BaSO4, which is a solid and falls out of solution.

So, initially, some of the Ba^2+ ions from BaCl2 react with SO4^2- ions from Na2SO4 to form BaSO4 precipitate. However, in the solution, there might be some unreacted Ba^2+ ions left.

Now, if more Na2SO4 solution is added, it will introduce more SO4^2- ions into the solution. These additional SO4^2- ions will continue to react with the remaining Ba^2+ ions to form more BaSO4 precipitate. As a result, more Ba^2+ ions will be removed from the solution as BaSO4.

Therefore, the addition of more Na2SO4 solution will decrease the concentration of Ba^2+ ions remaining in the solution. The correct answer is "Decreases the concentration of Ba^2+."