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Please answer the following question with explanation 

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Asked by Deepak 15th September 2018, 3:04 PM
Answered by Expert
Answer:
Λ = Equivalent cnductance (δ0)
 
 
Equivalent conductivity of BaCl= x1
 
Equivalent conductivity of H2SO4=x2
 
Equivalent conductivity of HCl =x3
 
δBaSO4 = δBaCl2 + δH2SO4 - δ 2HCl
 
            = δBaSO subscript 4 space end subscript space equals space fraction numerator 1000 cross times space Specific space conductance over denominator Solubility end fraction

straight x subscript 1 space plus space straight x subscript 2 space minus space 2 straight x subscript 3 space equals space fraction numerator 1000 straight y over denominator Solubility end fraction

Solubility space of space BaSO subscript 4 space end subscript space equals 1000 straight y space straight N

As space straight N space equals straight M over 2

Solubility space of space BaSO subscript 4 space end subscript space equals fraction numerator 1000 straight y space straight M over denominator 2 end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 1000 straight y space straight M over denominator 2 open parentheses straight x subscript 1 space plus space straight x subscript 2 space minus space 2 straight x subscript 3 close parentheses end fraction

BaSO subscript 4 space end subscript space rightwards arrow space Ba to the power of 2 plus end exponent space plus space SO subscript 4 to the power of 2 minus end exponent

straight k subscript sp subscript open parentheses BaSO subscript 4 close parentheses end subscript space end subscript space equals space open square brackets Ba to the power of 2 plus end exponent close square brackets space plus space open square brackets SO subscript 4 to the power of 2 minus end exponent close square brackets space straight M squared

space space space space space space space space space space space space space space equals fraction numerator 10 to the power of 6 straight y squared over denominator 4 open parentheses straight x subscript 1 space plus space straight x subscript 2 space minus space 2 straight x subscript 3 close parentheses squared end fraction
The solubility product of BaSO4 is  fraction numerator 10 to the power of 6 straight y squared over denominator 4 open parentheses straight x subscript 1 space plus space straight x subscript 2 space minus space 2 straight x subscript 3 close parentheses squared end fraction
Answered by Expert 19th September 2018, 12:18 PM
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