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Standard XII

Chemistry

Common Ion Effect

The solubilit...

Question

The solubility product $(K_{s p} m o l^{3} d m^{- 9})$ of $M X_{2}$ at 298 K based on the information available for the given concentration cell is (take $2.303 \times R \times 298 / F = 0.059 V$ ):

1 \times 10^{- 15}

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4 \times 10^{- 15}

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1 \times 10^{- 12}

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4 \times 10^{- 12}

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Solution

The correct option is B $4 \times 10^{- 15}$
$M / M^{2 +}$ (saturated solution of salt $M X_{2}) | | M^{2 +} (0.001 M)$ emf of concentration cell,
$E_{c e l l} = \frac{0.059}{n} l o g \frac{[M^{2 +}]_{R H S}}{[M^{2 +}]_{L H S}}$
$0.059 = \frac{0.059}{2} l o g \frac{[0.001]}{[M^{2 +}]}$
$∴ [M^{2 +}]_{L H S} = 10^{- 5} M$
Let solubility of salt be S mol/litre
thus $M X_{2} (S) \to M^{2 +} S + 2 X^{-} 2 S$
$∴ K_{s p} = 4 s^{3} = 4 \times (10^{- 5})^{3} = 4 \times 10^{- 15}$

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Similar questions

Q. The solubility product

(K_{s p} i n m o l^{3} d m^{- 9})

M X_{2}

298 K

based on the information available for the given concentration cell is: (take

2.303 \times R \times 298 / F = 0.059 V)

Q. The electrochemical cell shown below is a concentration cell,

M | M^{2 +}

(Saturated solution of a sparingly soluble salt,

M X_{2}) | | M^{2 +} (0.001 m o l d m^{- 3}) M .

The emf of the cell depends on the difference in concentration of

M^{2 +}

ions at the two electrodes. The emf of the cell at 298K is 0.059

The solubility product

(K_{s p} : m o l^{3} d m^{- 9})

M X_{2}

at 298 K based on the information available the given concentration cell is (take

2.303 \times R \times 298 / F = 0.059 V)

Q. The electrochemical cell shown below is a concentration cell,

M | M^{2 +}

(Saturated solution of a sparingly soluble salt,

M X_{2}) | | M^{2 +} (0.001 m o l d m^{- 3}) M .

The emf of the cell depends on the difference in concentration of

M^{2 +}

ions at the two electrodes. The emf of the cell at 298K is 0.059

The solubility product

(K_{s p} : m o l^{3} d m^{- 9})

M X_{2}

at 298 K based on the information available the given concentration cell is (take

2.303 \times R \times 298 / F = 0.059 V)

Q. Given the equilibrium constant,

K_{c}

of the reaction :

C u (s) + 2 A g^{+} (a q) \to C u^{2 +} (a q) + 2 A g (s)

10 \times 10^{15}

, calculate the

E_{c e l l}^{o}

of this reaction at

298 K

[2.303 \frac{R T}{F} a t 298 K = 0.059 V]

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:

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The solubility product (Kspmol3dm−9) of MX2 at 298 K based on the information available for the given concentration cell is (take 2.303×R×298/F=0.059V):

The correct option is B 4×10−15M/M2+ (saturated solution of salt MX2)||M2+(0.001M) emf of concentration cell,Ecell=0.059nlog[M2+]RHS[M2+]LHS0.059=0.0592log[0.001][M2+]∴[M2+]LHS=10−5MLet solubility of salt be S mol/litrethus MX2(S)→M2+S+2X−2S∴Ksp=4s3=4×(10−5)3=4×10−15

The solubility product $(K_{s p} m o l^{3} d m^{- 9})$ of $M X_{2}$ at 298 K based on the information available for the given concentration cell is (take $2.303 \times R \times 298 / F = 0.059 V$ ):