In: Chemistry
Question 4 – A sample has [Mg2+] = 60 mM and [SO42-] = 35 mM. There is an excess of solids containing MgSO4 at the bottom of the sample. Assume the sample is distilled water, is at 25 degrees Celsius, an under 1 atm of pressure. For reference, the thermodynamic equilibrium constant for MgSO4(s) dissolution at standard conditions is 9.5x103
4.a. Write the expression that would describe this system if it were at equilibrium.
4.b. Given the conditions above, what is the value of ion activity product (IAP) in this system? Neglect ion pairing and activity correction concerns.
4.c. Compare the IAP value with equilibrium constant. Would you predict more solid to dissolve or to precipitate? How many grams of MgSO4 would you expect to dissolve?
4.d. Assume the solution is allowed to equilibrate. What would the resulting activities of SO42- and Mg2+? Again assume no ion pairing and activity corrections?
4.e. Now let’s take the equilibrated system but assume the solution is seawater. How would the equilibrium constant change? How would the activities of SO42- and Mg2+ change? Would you expect more or less of the solid to dissolve?
4.f. Estimate the new equilibrium concentrations of SO42- and Mg2+
4 a ) MgSO4 (s) <---------------------> Mg+2 + SO4-2
equilibrium expression K = [Mg+2][SO42-]
b) [Mg2+] = 60 x 10^-3 M
[SO42-] = 35 x 10^-3 M.
Ionic product (IAP) = [Mg+2][SO42-]
= [60 x 10^-3][35 x 10^-3]
= 2.1 x 10^-3
c) equilibrium constant = 9.5 x 10^3
IAP vaue = 2.1 x 10^-3
IAP value < equilibrium constant
here solubility product is more so it is more dissolve. nearly 40 g will dissolve
d) MgSO4 (s) <---------------------> Mg+2 + SO4-2
S S
equilibrium expression K = [Mg+2][SO42-]
9.5 x 10^3 = S^2
S = 97.47 M
[Mg+2] = [SO42-] = 97.47 M
e)
In sea water more Mg+2 ions are available . so the concentration of Mg+2 ions increases. so the equibrium shifts backward direction .
less solid to dissolve