contestada

A 50.0 mL sample of 1.54×10−2 M NaSO4 is added to 50.0 mL of 1.28×10−2 Ca(NO3)2. What percentage of the Ca2+ remains unprecipitated?,

Respuesta :

The reaction would be as shown below;
Na2SO4 + Ca(NO3)2 = CaSO4 + 2 NaNO3
The moles of NaSO4 will be;
   = 0.05 × 0.0154 = 0.00077 moles
While the number of moles of Ca(NO3)2
   = 0.05 × 0.0128 = 0.00064 moles
The mole ratio of sodium sulfate and calcium nitrate is 1:1
Ca(NO3)2 is the limiting reactant, so ignoring the Ksp of CaSO4, zero percent of the Ca^2+ ions remain unprecipitated.