Consider the reactioncaso4(s)⇌ca2+(aq)+so2−4(aq)   at 25  ∘c  the equilibrium constant is  kc=2.4×10−5  for this reaction.- if excess caso4(s) is mixed with water at 25 ∘c to produce a saturated solution of caso4, what is the equilibrium concentration of so2−4? -  if the resulting solution has a volume of 1.2 l, what is the minimum mass of  caso4(s)  needed to achieve equilibrium?

a) According to the reaction equation:

CaSO4(s) ↔ Ca2+(aq)  + SO4-(aq)

when Kc = [Ca2+][SO4-]

when [Ca2+] = [SO4-] = X

and we have Kc = 2.4 x 10^-5 so ,by substitution we can get [Ca2+]&[SO4-] 

2.4 x 10^-5 = X^2

∴X = 0.0049

∴[SO4-] = [Ca2+] = 0.0049 M

b) when we have [Ca2+] =  0.0049 M so we can get the no.of moles of CaSO4:

moles of CaSO4 = molarity * volume 

                             = 0.0049 M * 1.2 L
                             = 0.00588 moles

when we know the molar mass of CaSO4 = 136.14 g/ mol, So we can get the mass:

∴mass of CaSO4 = moles of CaSO4 * molar mass of CaSO4

                             = 0.00588 moles * 136.14 g/mol

                             = 0.8 g

It idk haha what are we doing?