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