Asked by LT
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.
*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.
Answers
Answered by
DrBob222
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.
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.
Answered by
LT
BaSO4 less soluble so answer D
Answered by
DrBob222
Right, BaSO4 will be less soluble but it appears that is answer C, not D. You don't have the answers labeled.
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