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Equilibrium

Question
CBSEENCH11006621

If 25.0 cm3 of 0.50 M - Ba(NO3)are mixed with 25·0 cm3 of 0·0220M-NaF, will any BaF2 precipitate  [Ksp of Ba F2 is 1·7 × 10–6 at 298 K.]

Solution
Kspof BaF2 = 1·7 × 10–6 Total volume of the mixture solution = 25 + 25 = 50 cm3
left parenthesis straight i right parenthesis space Determining space of space Ba to the power of 2 plus end exponent space ion space concentration colon
left parenthesis Before space mixing right parenthesis space straight M subscript 1 straight V subscript 1 space equals space straight M subscript 2 straight V subscript 2 space left parenthesis After space mixing right parenthesis
space space space space space space space space space space space 0.05 space cross times space 25 space equals space straight M subscript 2 space cross times space 50
therefore space space space straight M subscript 2 left square bracket Molarity space of space Ba left parenthesis NO subscript 3 right parenthesis subscript 2 right square bracket space equals space fraction numerator 0.05 space cross times space 25 over denominator 50 end fraction space equals space 0.025 space straight M
Since space Ba left parenthesis NO right parenthesis subscript 3 space is space fully space ionised comma space
space space space space space space space space open square brackets Ba to the power of 2 plus end exponent close square brackets space equals space 0.025 space mol space ion space straight L to the power of negative 1 end exponent
left parenthesis ii right parenthesis space Determining space straight F to the power of minus space ion space concentration colon
left parenthesis Before space mixing right parenthesis space straight M subscript 1 straight V subscript 1 space equals space straight M subscript 2 straight V subscript 2 left parenthesis After space mixing right parenthesis
space space space space space space space space space space space 0.020 space cross times space 25 space space space space equals space straight M subscript 2 space cross times space 50
therefore space space space straight M subscript 2 left parenthesis Molarity space of space NaF right parenthesis space equals fraction numerator 0.020 space cross times space 25 over denominator 50 end fraction equals space 0.01 space straight M
As space NaF space is space fully space ionised
therefore space space space space space space space open square brackets straight F to the power of minus close square brackets space equals space 0.01 space mol space ions space straight L to the power of negative 1 end exponent
left parenthesis iii right parenthesis space Determining space ionic space product comma space
space space space space BaF subscript 2 space space rightwards harpoon over leftwards harpoon space space space Ba to the power of 2 plus end exponent space plus space 2 straight F to the power of minus
Ionic space product space equals space space open square brackets Ba to the power of 2 plus end exponent close square brackets space open square brackets straight F to the power of minus close square brackets squared
space space space space space space space space space space space space space space space space space space space equals left parenthesis 0.025 right parenthesis thin space left parenthesis 0.01 right parenthesis squared
space space space space space space space space space space space space space space space space space space space equals space 2.5 space cross times space 10 to the power of negative 6 end exponent
because space space Ionic space product space space equals space left parenthesis 2.5 space cross times space 10 to the power of negative 6 end exponent right parenthesis thin space greater than thin space straight K subscript sp space space left parenthesis 1.7 space cross times space 10 to the power of negative 6 end exponent right parenthesis
therefore space space space BaF subscript 2 space will space be space precipitated. space space space space space space space space space space