0

HETAV PATEL

Grade 11,

Courses New

The solubility product constant of Ca(OH)2 at25 degree celcius is 4.42*10^-5.A 500Ml of saturated solution of Ca(OH)2 is mixed with equal volume of 0.4M NaOH. What is mass of Ca(OH)2 precipitate out?

The solubility product constant of Ca(OH)2 at25 degree celcius is 4.42*10^-5.A 500Ml of saturated solution of Ca(OH)2 is mixed with equal volume of 0.4M NaOH. What is mass of Ca(OH)2 precipitate out?
 
 
 

Grade:12

1 Answers

Vikas TU
14149 Points
6 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)

Think You Can Provide A Better Answer ?

Other Related Questions on 9 grade maths

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More