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Home / Doubts and Solutions/NEET/Chemistry

On adding Na2SO4 to 0.1 M solution each of Ag+, Ba+2, Ca+2 ions the substance first precipitated is (given KspBaSO4 = 10–11, KspCaSO4 = 10–6, KspAg2SO4 = 10–5)

Asked by Atulcaald 15th May 2018, 1:06 PM
Answered by Expert
Answer:
F o r space A g subscript 2 S O subscript 4 comma space K s p space equals space left square bracket A g to the power of plus right square bracket squared space left square bracket S O subscript 4 to the power of 2 minus end exponent right square bracket therefore space left square bracket S O subscript 4 to the power of 2 minus end exponent right square bracket space n e e d e d space f o r space p r e c i p i t a t i o n space o f space A g subscript 2 S O subscript 4 comma greater than space fraction numerator K s p over denominator left square bracket A g to the power of plus right square bracket squared end fraction space equals space fraction numerator 10 to the power of negative 5 end exponent over denominator 0.1 space cross times 0.1 end fraction space equals space 10 to the power of negative 3 end exponent M F o r space C a S O subscript 4 comma space K s p space equals space left square bracket C a to the power of 2 plus end exponent right square bracket space space left square bracket S O subscript 4 to the power of 2 minus end exponent right square bracket therefore space left square bracket S O subscript 4 to the power of 2 minus end exponent right square bracket space n e e d e d space f o r space p r e c i p i t a t i o n space o f space C a S O subscript 4 comma space space greater than space fraction numerator K s p over denominator left square bracket C a to the power of 2 plus end exponent right square bracket end fraction space equals space fraction numerator 10 to the power of negative 6 end exponent over denominator 0.1 end fraction space equals space 10 to the power of negative 5 end exponent M F o r space B a S O subscript 4 comma space K s p space equals space left square bracket B a to the power of 2 plus end exponent right square bracket space space left square bracket S O subscript 4 to the power of 2 minus end exponent right square bracket therefore space left square bracket S O subscript 4 to the power of 2 minus end exponent right square bracket space n e e d e d space f o r space p r e c i p i t a t i o n space o f B a S O subscript 4 space end subscript greater than space space fraction numerator K s p over denominator space left square bracket B a to the power of 2 plus end exponent right square bracket end fraction space space equals space space fraction numerator 10 to the power of negative 11 end exponent over denominator 0.1 end fraction space equals space 10 to the power of negative 10 end exponent M S O subscript 4 to the power of 2 minus end exponent space space greater than space 10 to the power of negative 10 end exponent M T h u s comma space m i n i m u m space space left square bracket S O subscript 4 to the power of 2 minus end exponent right square bracket space i s space r e q u i r e d space f o r space p r e c i p i t a t i o n space o f space B a S O subscript 4 space a n d space h e n c e space i t space i s space p r e c i p i t a t e space o u t space f i r s t.
Answered by Expert 16th May 2018, 12:32 PM
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