You visited us 1 times! Enjoying our articles? Unlock Full Access!

Common Ion Effect

An excess of ...

Question

An excess of $A g_{2} C r O_{4} (s)$ is added to a $5 \times 10^{- 3} M$ $K_{2} C r O_{4}$ solution. The concentration of $A g^{+}$ in the solution is closest to.[Solubility product for $A g_{2} C r O_{4} = 1.1 \times 10^{- 12}$ ]

2.2 \times 10^{- 10}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.5 \times 10^{- 5}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.0 \times 10^{- 6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.0 \times 10^{- 3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $1.5 \times 10^{- 5}$ M
Since, $K_{2} C r O_{4}$ is a strong electrolyte so the concentration of chromate ion will be almost equal to the concentration of $K_{2} C r O_{4}$ .

$A g_{2} C r O_{4} ⇋ 2 A g^{+} + C r O_{4}^{2 -}$
$S$ $2 S$ $5 \times 10^{- 3} M$

Now,
$K_{s p} = [A g^{+}]^{2} [C r O_{4}^{2 -}]$

$1.1 \times 10^{- 12} = [A g^{+}]^{2} [5 \times 10^{- 3}]$

$[A g^{+}] = 1.5 \times 10^{- 5}$

Suggest Corrections

Similar questions

The $K_{s p}$ of $A g_{2} C r O_{4}$ is $1.1 \times 10^{- 12}$ at 298K. The solubility (in mol/L) of $A g_{2} C r O_{4}$ in 0.1M $A g N O_{3}$ solution is:

Q. The

K_{s p} o f A g_{2} C r O_{4} i s 1.1 \times 10^{- 12} a t 298 K .

The solubility (in

m o l / L

) of

A g_{2} C r O_{4}

in a

0.1 M A g N O_{3}

solution is

Q. The

K_{s p} o f A g_{2} C r O_{4} i s 1.1 \times 10^{- 12} a t 298 K .

The solubility (in

m o l / L

) of

A g_{2} C r O_{4}

in a

0.1 M A g N O_{3}

solution is

Q. The

K_{s p}

A g_{2} C r O_{4}

1.1 \times 10^{- 12}

at 298 K. The solubility (in mole / L) of

A g_{2} C r O_{4}

in a

0.1 M A g N O_{3}

solution is

Q. Solid

A g N O_{3}

is added to a solution which is

0.1 M

C l^{-}

and

0.1 M

C r O_{4}^{2 -}

. If

K_{s p}

values for

A g C l

and

A g_{2} C r O_{4}

solutions are

1.7 \times 10^{- 10}

and

1.9 \times 10^{- 12}

respectively. Find the minimum concentration of

C l^{-}

when

A g_{2} C r O_{4}

starts precipitating.

Join BYJU'S Learning Program

An excess of Ag2CrO4(s) is added to a 5×10−3M K2CrO4 solution. The concentration of Ag+ in the solution is closest to.[Solubility product for Ag2CrO4=1.1×10−12]

The correct option is B 1.5×10−5MSince, K2CrO4 is a strong electrolyte so the concentration of chromate ion will be almost equal to the concentration of K2CrO4.Ag2CrO4⇋2Ag++CrO2−4S 2S 5×10−3 MNow,Ksp=[Ag+]2[CrO2−4]1.1×10−12=[Ag+]2[5×10−3][Ag+]=1.5×10−5

An excess of $A g_{2} C r O_{4} (s)$ is added to a $5 \times 10^{- 3} M$ $K_{2} C r O_{4}$ solution. The concentration of $A g^{+}$ in the solution is closest to.[Solubility product for $A g_{2} C r O_{4} = 1.1 \times 10^{- 12}$ ]