JEE Main 24 Feb 2021 Shift 1 Solved Paper
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The stepwise formation of [ C u ( N H 3 ) 4 ] 2 + is given below
C u 2 + + N H 3 ⇌ [ C u ( N H 3 ) ] 2 +
[ C u ( N H 3 ) ] 2 + + N H 3
[ C u ( N H 3 ) 2 ] 2 +
[ C u ( N H 3 ) 2 ] 2 + + N H 3
[ C u ( N H 3 ) ] 2 +
[ C u ( N H 3 ) 3 ] 2 + + N H 3
[ C u ( N H 3 ) 4 ] 2 +
The value of stability constantsK 1 , K 2 , K 3 and K 4 are 10 4 , 1.58 × 10 3 , 5 × 10 2 and 10 2 respectively.
The overall equilibrium constants for dissociation of[ C u ( N H 3 ) ] 2 + ] is x × 10 − 12 . The value of x is ............. . (Rounded off to the nearest integer)
The value of stability constants
The overall equilibrium constants for dissociation of
[24 Feb 2021 Shift 1]
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