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Ionic Equilibrum question

2013 · Q113
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Ionic Equilibrum question

2013 · Q113

NEETChemistryIonic EquilibrumMCQ+4 / −1
The values of Ksp of CaCO3 and CaC2O4 are 4.7 ×\times× 10−-−9 and 1.3 ×\times× −-−9 respectively at 25oC. If the mixture of these two is washed with water, what is the concentration of Ca2+ ions in water ?
  1. A
    5.831 ×\times× 10−-−5 M
  2. B
    6.856 ×\times× 10−-−5 M
  3. C
    3.606 ×\times× 10−-−5 M
  4. D
    7.746 ×\times× 10−-−5 M
View written solutionFree

Correct answer: D

CaCO3$ \to $Ca2++CO32-
xx

.tg {border-collapse:collapse;border-spacing:0;} .tg td{font-family:Arial, sans-serif;font-size:14px;padding:10px 5px;border-style:solid;border-width:0px;overflow:hidden;word-break:normal;border-top-width:1px;border-bottom-width:1px;border-color:black;} .tg th{font-family:Arial, sans-serif;font-size:14px;font-weight:normal;padding:10px 5px;border-style:solid;border-width:0px;overflow:hidden;word-break:normal;border-top-width:1px;border-bottom-width:1px;border-color:black;} .tg .tg-9wq8{border-color:inherit;text-align:center;vertical-align:middle} .tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top}
CaC2O4$ \to $Ca2++C2O42-
yy


[Ca2+] = x + y

Now, Ksp (CaCO3) = [Ca2+] [CO3 2-]

or 4.7 × 10–9 = (x + y)x .......(1)

similarly, Ksp (CaC2O4) = [Ca2+] [C2O42–]

or 1.3 × 10–9 = (x + y)y .......(2)

Dividing equation (1) and (2), we get

${x \over y} = 3.6$

$ \therefore $ x = 3.6y

Putting in equation (2) we get

y(3.6y + y) = 1.3 × 10–9

$ \Rightarrow $ y = 1.68 $ \times $ 10-5

and x = 3.6 $ \times $ 1.68 $ \times $ 10-5 = 6.048 $ \times $ 10-5

$ \therefore $ [Ca2+] = (x + y) = (6.048 $ \times $ 10-5) + (1.68 $ \times $ 10-5)

$ \therefore $ [Ca2+] = 7.746 × 10–5 M
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