Calculate the solubility and solubility product of Co2 - FeCN6 - in water at 25-C from the following data -Conductivity of a structured solution of Co2 - FeCN6 - is 2-06- 10-6- -1cm2mol-1 and 4-1- 10-7- -1cm-1- The ionic molar conductivities of Co2- and FeCN64- are 86-0- -1cm-1 and 444-0- -1cm-1mole-1-then the Ksp is-

The correct option is D 7.682× 10^ 17 λ _M=k× 100/M (86× 2+444)=(2.06× 10^ 6 4.1× 10^ 7)× 1000/s s=2.678× 10^ 6 K_sp=4s^3=7.687× 10^ 1 ...

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Standard XII

Chemistry

Conductivity of Ionic Solutions

Calculate the...

Question

Calculate the solubility and solubility product of $C o_{2} [F e (C N)_{6}]$ in water at $25^{\circ} C$ from the following data :
Conductivity of a structured solution of $C o_{2} [F e (C N)_{6}]$ is $2.06 \times 10^{- 6} Ω^{- 1} c m^{2} m o l^{- 1}$ and $4.1 \times 10^{- 7} Ω^{- 1} c m^{- 1}$ . The ionic molar conductivities of $C o^{2 +}$ and $F e (C N)_{6}^{4 -}$ are $86.0 Ω^{- 1} c m^{- 1}$ and $444.0 Ω^{- 1} c m^{- 1} m o l e^{- 1}$ .,then the $K_{s p}$ is:

7.682 \times 10^{- 14}

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7.682 \times 10^{- 18}

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3.841 \times 10^{- 18}

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7.682 \times 10^{- 17}

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Solution

The correct option is D $7.682 \times 10^{- 17}$
$λ_{M} = \frac{k \times 1000}{M}$
$(86 \times 2 + 444) = \frac{(2.06 \times 10^{- 6} - 4.1 \times 10^{- 7}) \times 1000}{s}$
$s = 2.678 \times 10^{- 6}$
$K_{s p} = 4 s^{3} = 7.687 \times 10^{- 17}$

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Similar questions

Q. Which of the following options are correct for solubility and solubility product of

C O_{2} [F e {(C N)}_{6}]

in water at

25^{\circ}

. Conductivity of a saturated solution of

C O_{2} [F e {(C N)}_{6}]

2.06 \times 10^{- 6} Ω^{- 1} {cm}^{- 1}

and that of water used is

4.1 \times 10^{- 7} Ω^{- 1} {cm}^{- 1} .

The ionic molar conductivities of

C o^{2 +} and F e (C N)_{6}^{4 -}

are

86.0 Ω^{- 1} {cm}^{2} {mol}^{- 1} and 444.0 Ω^{- 1} {cm}^{2} {mol}^{- 1},

respectively.

Q. The specific conductance of a saturated solution of

A g C l

25^{\circ} C

3.55 \times 10^{- 6} Ω^{- 1} c m^{- 1}

and that of water used for preparing the solution is

1.60 \times 10^{- 6} Ω^{- 1} c m^{- 1}

. What is the solubility product of AgCl?

(Λ^{\infty} (A g C l) = 150.0 Ω^{- 1} c m^{2} {eqv}^{- 1})

Q. Conductivity of a saturated solution of

C u_{2} [F e (C N)_{6}]

after subtracting the conductivity of water is

1.28 \times 10^{- 5} Ω^{- 1} c m^{- 1}

.
Calculate value of solubillity of

C u [F e (C N)_{6}]

[Λ_{m}^{x} (C u S O_{4}) = 260 S c m^{2} m o l^{- 1}

Λ_{m}^{x} (K_{2} S O_{4}) = 300 S c m^{2} m o l^{- 1}

Λ_{m}^{x} (K_{4} F e (C N)_{6}) = 720 S c m^{2} m o l^{- 1}]

Report your answer as (solubility)

\times (10^{5})

Q. The specific conductance of a saturated solution of AgCl is found to be

1.86 \times 10^{- 6} o h m^{- 1} c m^{- 1}

and that of water is

6 \times 10^{- 8} o h m^{- 1} c m^{- 1}

. The solubility of AgCl given that

λ_{A}^{0} g C l = (137.2 o h m^{- 1} c m^{- 1}

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The specific conductance of $A g C l$ at 298 K is $1.826 \times 10^{- 6} o h m^{- 1} c m^{- 1}$ . The ionic conductivities of $A g^{+}$ and $C l^{-}$ ions are $61.92$ and $76.34$ $o h m^{- 1} c m^{2} m o l^{- 1}$ respectively. What is the solubility of $A g C l$ in water? (Molar mass of $A g C l$ is $143.5 g m o l^{- 1}$ )

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The correct option is D 7.682×10−17λM=k×1000M(86×2+444)=(2.06×10−6−4.1×10−7)×1000ss=2.678×10−6Ksp=4s3=7.687×10−17

The correct option is D $7.682 \times 10^{- 17}$
$λ_{M} = \frac{k \times 1000}{M}$
$(86 \times 2 + 444) = \frac{(2.06 \times 10^{- 6} - 4.1 \times 10^{- 7}) \times 1000}{s}$
$s = 2.678 \times 10^{- 6}$
$K_{s p} = 4 s^{3} = 7.687 \times 10^{- 17}$