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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > when the pH of a saturated solution of M(OH)2 is 13....
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when the pH of a saturated solution of M(OH)2 is 13. Hence Ksp of M(OH)2 is

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POH = 1 2OH = 10^-1 OH = 0.05 M s = 0.05 M Ksp = 4s^3 Ksp = 5 x 10^-4 M^3
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M(OH)2 gives M+(ion) + 2 (OH)- ions. So concentration of M+ ion is half that of OH- ions. To calculate the concentration of OH minus ions we need pOH value which is equal to {14 - pH}. In this question pH is 13 therefore pOH = 1. Concentration of OH ions is given by = 10^ (-pOH). ...
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M(OH)2 gives M+(ion) + 2 (OH)- ions. So concentration of M+ ion is half that of OH- ions. To calculate the concentration of OH minus ions we need pOH value which is equal to {14 - pH}. In this question pH is 13 therefore pOH = 1. Concentration of OH ions is given by [OH] = 10^ (-pOH). = 10^-1 =0.1 M (here M is molarity, mol/l) To calculate concentration of metal ion namely [M] is half of the concentration of OH ion =(0.1)/2 =0.05 M once we have concentration of these let us now calculate the Ksp value. Ksp for this = [M] * [OH]^2 = 0.05 * (0.1)^2 =0.0005 = 5* !0^(-4) units will be M^3 read less
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Specialization in Organic Chemistry. Teaching Chemistry has always been my passion.

5 X 10-4 or 0.0005
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IIT qualified chemist

0.004
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Chemistry expert

pH=13 pOH=1 =0.1 =2S Ksp=4s3=5 X 10-4
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pH value = 13 , pOH = 14-13 = 1 , = 10-1 = 0.1 , = 0.1/2 = 0.05 , Ksp = 2 = (0.05)2(0.1) Ksp = 0.0025 x 0.1 = 2.5x10-4
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pH = 13 , pOH = 14-13 = 1 , = 10-1 = 0.1 , = 0.1/2 = 0.05 Ksp = 2 = 0.1 x (.05)2 = 0.1x0.0025 = 2.5x10-4
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IIT/BITSAT Decoded !!

Hi K.kamal Here is the solution to your problem: M(OH)2 ---> M2+ + 2OH- (1) Ksp = X^2 (2) Now concentration of OH- can be calculated as pOH+PH =14 at 25C taken as standard. therefore POH=1 { ph=13 given } hence OH- =10^-1 Now conc. of M2+ will be half...
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Hi K.kamal Here is the solution to your problem: M(OH)2 ---> M2+ + 2OH- (1) Ksp = [M2+]X[OH-]^2 (2) Now concentration of OH- can be calculated as pOH+PH =14 at 25C taken as standard. therefore POH=1 { ph=13 given } hence OH- =10^-1 Now conc. of M2+ will be half the conc. of OH- at equib. as it is clear from above equation(1) therefore putting values in (2) Ksp= (0.5x10^-1)x(10^-1)^2 = 0.5x10^-3 mol^3litre^-3 read less
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Professional Chemistry Teacher for IIT JEE/NEET UG,Online expert for IB/IGCSE/AP CURRICULUM

M(OH)2 is a base,so PH+POH=14 POH=1,=0.1Normal 2=0.1 =0.1/2=0.05normal For M(OH)2,Ksp=4SPower 3 Ksp=0.0005
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Math Educator for Std.11th ,12th , Engineering Entrance and Degree Level with 11+ Years Experience

PLEASE GO THROUGH INSERT EQUATION OPTION IN MS WORD. YOU WILL GET MY ANSWER IN CLEAR FORMAT ?M(OH)?_2 ? M^(2+ )+ 2 ?OH?^- Ksp= ^2 P^H = 13 P^OH = 14 -- 13 = 1 = ?10?^(-P^OH...
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PLEASE GO THROUGH INSERT EQUATION OPTION IN MS WORD. YOU WILL GET MY ANSWER IN CLEAR FORMAT ?M(OH)?_2 ? M^(2+ )+ 2 ?OH?^- Ksp= [M^(2+) ] [?OH?^- ]^2 P^H = 13 P^OH = 14 – 13 = 1 [?OH?^- ] = ?10?^(-P^OH ) = ?10?^(-1) M [M^(2+) ]= ?10?^(-1)/2 = 5.0 × ?10?^(-2) Ksp=(5.0 × ?10?^(-2) )?(?10?^(-1))?2 Ksp of M(OH)2 is = 5.0 × 10^(-4) read less
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