AgNO3 solution is added to the solution containing equal moles of NaCl, NaBr, NaI and Na2 CrO4 the Ksp of Ag2CrO4 , AgCl, AgBr and AgI are respectively, 1.1 × 10-12 , 1.8 × 10-10 , 5.0 × 10-13, 8.3 × 10-17. Which one of the following salts will precipitate at the last?

Question

AgNO3 solution is added to the solution containing equal moles of NaCl, NaBr, NaI and Na2 CrO4 the Ksp of Ag2CrO4 , AgCl, AgBr and AgI are respectively, 1.1 × 10-12 , 1.8 × 10-10 , 5.0 × 10-13, 8.3 × 10-17. Which one of the following salts will precipitate at the last? (A) AgCl  (B) Ag2CrO4  (C) AgI (D) AgBr

Tardigrade Admin · Accepted Answer

Salt Ksp Solublity Ag2CrO4 1.1 × 10-12 = 4s3 s = √[3](Ksp/4) = 0.65 × 10-4 AgCl 1.8 × 10-10 = s2 s = √Ksp = 1.34 × 10-5 AgBr 5 × 10-13 = s2 s = √Ksp = 0.71 × 10-6 AgI 8.3 × 10-17 = s2 s = √Ksp = 0.9 × 10-8 Solubility of Ag2 CrO4 is highest thus, it will be precipitated at last.

Salt	$K_{s p}$	Solublity
$A g_{2} C r O_{4}$	$1.1 \times 1 0^{- 12} = 4 s^{3}$	$s = 3 \frac{K _{s p}}{4} = 0.65 \times 1 0^{- 4}$
$A g Cl$	$1.8 \times 1 0^{- 10} = s^{2}$	$s = K_{s p} = 1.34 \times 1 0^{- 5}$
$A g B r$	$5 \times 1 0^{- 13} = s^{2}$	$s = K_{s p} = 0.71 \times 1 0^{- 6}$
$A g I$	$8.3 \times 1 0^{- 17} = s^{2}$	$s = K_{s p} = 0.9 \times 1 0^{- 8}$

Q. AgNO3​ solution is added to the solution containing equal moles of NaCl,NaBr,NaI and Na2​CrO4​ the Ksp of Ag2​CrO4​,AgCl,AgBr and AgI are respectively, 1.1×10−12,1.8×10−10,5.0×10−13,8.3×10−17. Which one of the following salts will precipitate at the last?

AgCl

Ag2​CrO4​

AgI

AgBr

Salt Ksp​ Solublity Ag2​CrO4​ 1.1×10−12=4s3 s=34Ksp​​​=0.65×10−4 AgCl 1.8×10−10=s2 s=Ksp​​=1.34×10−5 AgBr 5×10−13=s2 s=Ksp​​=0.71×10−6 AgI 8.3×10−17=s2 s=Ksp​​=0.9×10−8 Solubility of Ag2​CrO4​ is highest thus, it will be precipitated at last.

$A g Cl$

$A g_{2} C r O_{4}$

$A g I$

$A g B r$