Vikas TU
Last Activity: 7 Years ago
Dear Student,
From the question:
ionisation eqn. of Ca(Oh)2: Ca(OH)2 = Ca2+ + 2OH–
ksp of Ca(OH)2=[Ca 2+] [OH-]2=s*(2s)^2=4s^3
=>s=cube-root(ksp/4)=sqrt(4.42*10^-5 /4)=2.23*10^-2 M
Ca 2+ ions in 500 ml soln.=(2.23*10^-2)/2 =0.011
500 ml soln+500 ml NaOH=1000ml,molarity of OH- in new solution=0.2M
[Ca]2+ =ksp/[OH-]^2 = (4.42*10^-5)/(0.2*0.2)=0.0011M
no. of moles of Ca2+ ppt=0.011-0.0011=0.0102M. Mass of Ca(OH)2= 0.0102*74.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
