Asked by Anonymous
Consider the following reaction.
CaSO4(s) reverse reaction arrow Ca2+(aq) + SO42-(aq)
At 25°C the equilibrium constant is Kc = 2.4x10^-5 for this reaction.
(a) If excess CaSO4(s) is mixed with water at 25°C to produce a saturated solution of CaSO4, what are the equilibrium concentrations of Ca2+ and SO42-?
(b) If the resulting solution has a volume of 1.4 L, what is the minimum mass of CaSO4(s) needed to achieve equilibrium?
For a I got 4.9*10^5 M for both concentrations. I'm not sure how to do b. Can someone show me what steps to do?
CaSO4(s) reverse reaction arrow Ca2+(aq) + SO42-(aq)
At 25°C the equilibrium constant is Kc = 2.4x10^-5 for this reaction.
(a) If excess CaSO4(s) is mixed with water at 25°C to produce a saturated solution of CaSO4, what are the equilibrium concentrations of Ca2+ and SO42-?
(b) If the resulting solution has a volume of 1.4 L, what is the minimum mass of CaSO4(s) needed to achieve equilibrium?
For a I got 4.9*10^5 M for both concentrations. I'm not sure how to do b. Can someone show me what steps to do?
Answers
Answered by
phil
.005
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