pOH=14-pH=-lg[OH^-]
[OH^-]=-10^(-(14-11.64))=4.37*10^(-3) mol/L
Ca(OH)2=Ca^(2+)+2OH^-
[Ca^(2+)]=4.37*10^(-3)/2=2.18*10^(-3) mol/L
mol Ca^(2+)=2.55*2.18*10^(-3)=5.57*10^(-3)
m=5.57*10^(-3)mol*40.078g/mol=0.223g
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