In a saturated solution of A g B r ( K s p = 2 × 10 - 13 ) , if 10 - 7 moles of A g N O 3 are added to 1 litre of this solution, find specific conductance of this solution in terms of 10 - 7 ? S ? m - 1 units.

Question

In a saturated solution of $A g B r (K_{s p} = 2 \times 10^{- 13})$ , if $10^{- 7}$ moles of $A g N O_{3}$ are added to 1 litre of this solution, find specific conductance of this solution in terms of $10^{- 7} S m^{- 1}$ units.

Given: $λ_{(A g^{+})}^{\circ} = 6 \times 10^{- 3} S m^{2} / m o l$

$λ_{(N O_{3}^{-})}^{\circ} = 7 \times 10^{- 3} S m^{2} / m o l$

$λ_{(B r^{-})}^{\circ} = 8 \times 10^{- 3} S m^{2} / m o l$

Smart Achievers · Accepted Answer

(69)

$A g B r_{(s)} ⇌ \underset{(s + 10^{- 7})}{A g^{+}} + \underset{s}{B r^{-}}$

$s^{2} + 10^{- 7} s - 20 \times 10^{- 14} = 0$

$s = \frac{- 10^{- 7} + \sqrt{10^{- 14} + 80 \times 10^{- 14}}}{2} = 4 \times 10^{- 7}$

$[A g^{+}] = 5 \times 10^{- 7} M$

$[B r^{-}] = 4 \times 10^{- 7} M$

$[N O_{3}^{-}] = 1 \times 10^{- 7} M$

$Λ ° = \frac{K}{1000 C}$ $\Rightarrow K_{A g^{+}}^{1} = 6 \times 10^{- 3} \times 1000 \times 5 \times 10^{- 7}$

$= 30 \times 10^{- 7} s^{- 1} m^{2} / m o l$

$K_{N O_{3}^{-}}^{1} = 7 \times 1 \times 10^{- 7} s m^{2} / m o l$

$K_{B r^{-}}^{1} = 8 \times 4 \times 10^{- 7} s m^{2} / m o l = 32 \times 10^{- 7}$

$K_{total}^{1} = 69 \times 10^{- 7} s m^{2} / m o l$

Solution

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Solution

Q. In a saturated solution of AgBr(Ksp=2×10-13), if 10-7 moles of AgNO3 are added to 1 litre of this solution, find specific conductance of this solution in terms of 10-7 S m-1 units. Given: λ(Ag+)∘=6×10-3 S m2/mol λ(NO3-)∘=7×10-3 S m2/mol λ(Br-)∘=8×10-3 S m2/mol

Ans. (69) AgBr(s)⇌Ag+(s+10-7)+Br-s s2+10-7s-20×10-14=0 s=-10-7+10-14+80×10-142=4×10-7 [Ag+]=5×10-7 M [Br-]=4×10-7 M [NO3-]=1×10-7 M Λ°=K1000 C⇒KAg+1=6×10-3×1000×5×10-7 =30×10-7 s-1 m2/mol KNO3-1=7×1×10-7 s m2/mol KBr-1=8×4×10-7 s m2/mol=32×10-7 Ktotal1=69×10-7 sm2/mol

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