Question
CBSEENCH11008267

pH of a saturated solution of Ba(OH)2 is 12. The value of solubility product Ksp of Ba(OH)2 is

  • 3.3 x 10-7

  • 5.0 x 10-7

  • 4.0 x 10-6

  • 5.0 x 10-6

Solution

B.

5.0 x 10-7

Given, pH of Ba(OH)2 = 12
pOH = 14-pH
= 14-12 = 2
We know that,
pOH = -log [OH-]
2 =-log [OH-]
[OH-] = antilog (-2)
[OH-] = 1 x 10-2
Ba(OH)2dissolves in water as 
stack Ba left parenthesis OH right parenthesis subscript 2 space left parenthesis straight s right parenthesis with straight s space mol space straight L to the power of negative 1 end exponent below space rightwards harpoon over leftwards harpoon space stack Ba to the power of 2 plus end exponent with straight s below space plus stack 2 OH to the power of minus with 2 straight s below
left square bracket OH to the power of minus right square bracket space equals space 2 straight s space equals space 1 space straight x space 10 to the power of negative 2 end exponent
left square bracket Ba to the power of 2 plus end exponent right square bracket space equals space fraction numerator left square bracket OH to the power of minus right square bracket over denominator 2 end fraction space equals space fraction numerator 1 space straight x space 10 to the power of negative 2 end exponent over denominator 2 end fraction
straight K subscript sp space equals space left square bracket Ba to the power of 2 plus end exponent right square bracket left square bracket OH to the power of minus right square bracket squared
space equals space open parentheses fraction numerator 1 space straight x space 10 to the power of negative 2 end exponent over denominator 2 end fraction close parentheses left parenthesis 1 straight x 10 to the power of negative 2 end exponent right parenthesis squared
space equals space 0.5 space straight x space 10 to the power of negative 6 end exponent space equals space 5 space straight x space 10 to the power of negative 7 end exponent

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