Answer to Question #169504 in General Chemistry for fatima

Question

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

Answer 1

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