Ca(OH)2 --> Ca2+ + 2OH-;
pOH=14 - pH=14-11.6=2.4;
pOH=-lg[OH-];
[OH-]=10-pOH=10-2.4=0.00398≈4*10^-3 mol/L;
[Ca2+]=[OH-]/2=2*10^-3 mol/L;
[SO42-]=[Ca2+]=2*10^-3
Answer: 2*10^-3