Answer to Question #169504 in General Chemistry for fatima

Question #169504

ph of a saturated solution of ba(oh)2 is 12. the value of solubility of product (ksp) of ba(oh)2 is


1
Expert's answer
2021-03-08T06:00:31-0500

Solution:


Ba(OH)2⇄Ba2+2OH


pH=12⇒p(OH)=14−pH → p(OH)=14−12=2


[OH]=10−pOH=10−2 or 1×10−2



as the concentration of Ba2+ is half of OH

 


→Ba2+=0.5×10−2

 


Ksp​=(0.5×10−2)(1×10−2)2



Ksp = 0.5×10−6 = 5×10−7



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