If equal volumes of and XY (both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of at 300 K? (Given (at 300 K) for ) [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
If equal volumes of $A B_{2}$ and XY (both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of $A Y_{2}$ at 300 K

(Given $K_{s p}$ (at 300 K) for $A Y_{2} = 5.2 \times 10^{- 7}$ ) [2025]

1 $3.6 \times 10^{- 3} {M AB}_{2}, 5.0 \times 10^{- 4} M XY$

2 $2.0 \times 10^{- 4} {M AB}_{2}, 0.8 \times 10^{- 3} M XY$

3 $2.0 \times 10^{- 2} {M AB}_{2}, 2.0 \times 10^{- 2} M XY$

4 $1.5 \times 10^{- 4} {M AB}_{2}, 1.5 \times 10^{- 3} M XY$

Ans.
(3)

$A Y_{2} (s) ⇌ A^{2 +} (a q) + 2 Y^{-} (a q)$

$Q_{s p} (A Y_{2}) = [A^{2 +}] {[Y^{-}]}^{2}$

By mixing equal volumes of two solutions, concentration of each is reduced to half.

Precipitation occurs when $Q_{s p} (A Y_{2}) > K_{s p} (A Y_{2})$

(1)   $Q_{s p} = [A^{2 +}] {[Y^{-}]}^{2}$

$= (\frac{3.6 \times 10^{- 3}}{2}) \times {(\frac{5.0 \times 10^{- 4}}{2})}^{2}$

$= 1.125 \times 10^{- 10} < K_{s p} (5.2 \times 10^{- 7})$ (no ppt)

(2)   $Q_{s p} = [A^{2 +}] {[Y^{-}]}^{2}$

$= (\frac{2.0 \times 10^{- 4}}{2}) \times {(\frac{0.8 \times 10^{- 3}}{2})}^{2}$

   $= 1.6 \times 10^{- 11} < K_{s p} (5.2 \times 10^{- 7})$ (no ppt)

(3)   $Q_{s p} = [A^{2 +}] {[Y^{-}]}^{2}$

$= (\frac{2.0 \times 10^{- 2}}{2}) \times {(\frac{2.0 \times 10^{- 2}}{2})}^{2}$

   $= 1.0 \times 10^{- 6} > K_{s p} (5.2 \times 10^{- 7})$ (ppt formed)

(4)   $Q_{s p} = [A^{2 +}] {[Y^{-}]}^{2}$

$= (\frac{1.5 \times 10^{- 4}}{2}) \times {(\frac{1.5 \times 10^{- 3}}{2})}^{2}$

$= 4.21875 \times 10^{- 11} < K_{s p} (5.2 \times 10^{- 7})$ (no ppt)

Want Us to Email you About Special Offers & Updates?