Question 10
a 75.0 −ml sample of 1.54×10−2 m na2so4(aq) is added to 75.0 ml of 1.22×10−2 m ca(no3)2(aq).
part a
what percentage of the ca2+ remains unprecipitated?
express your answer using two significant figures.

The reaction that should be followed is
Na2SO4 + Ca(NO3)2 --> CaSO4 + 2NaNO3
first calculate the limiting reactant
mol Na2SO4 = 0.075 L (1.54×10−2 mol / L) = 1.155x10-3 mol

mol Ca(NO3)2 = 0.075 L (1.22×10−2 mol / L) = 9.15x10-4 mol
so the limiting reactant is the Ca(NO3)2

so all of the Ca2+ will be precipitated, percentage unprecipitated = 0.00 % 

Tolerance to high levels of salinity

Ithink that the answer is b