If the solubility of AgCl formula mass -143 in water at 25oC is 1-43- 10-4 g -100 mL of solution then the value of Ksp will be-

The correct option is A 1× 10^ 5 AgCl → Ag^++Cl^ K_sp=[Ag^+][Cl^ ] Concentration = 1.43×10^ 4/143mol/100 ml #160; #160; #160; # ...

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Byju's Answer

Standard XII

Chemistry

Solubility Product

If the solubi...

Question

If the solubility of $A g C l$ (formula mass $= 143$ ) in water at $25^{o}$ C is $1.43 \times 10^{- 4}$ g $/ 100$ mL of solution then the value of $K_{s p}$ will be:

1 \times 10^{- 5}

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2 \times 10^{- 5}

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1 \times 10^{- 10}

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2 \times 10^{- 10}

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Solution

The correct option is A $1 \times 10^{- 5}$
$A g C l \to A g^{+} + C l^{-}$
$K_{s p} = [A g^{+}] [C l^{-}]$

Concentration = $\frac{1.43 \times 10^{- 4}}{143} m o l / 100 m l$

= $10^{- 6} m o l / 100 m l$

= $\frac{10^{- 6} \times 1000}{100} m o l / L i t e r$

= $10^{- 5} m o l / L$
$K_{s p} = 10^{- 5} \times 10^{- 5}$
$= 10^{- 10}$

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If the solubility of AgCl (formula mass =143) in water at 25oC is 1.43×10−4 g/100 mL of solution then the value of Ksp will be:

The correct option is A 1×10−5AgCl→Ag++Cl−Ksp=[Ag+][Cl−]Concentration =1.43×10−4143mol/100ml =10−6mol/100ml =10−6×1000100mol/Liter = 10−5mol/LKsp=10−5×10−5=10−10

If the solubility of $A g C l$ (formula mass $= 143$ ) in water at $25^{o}$ C is $1.43 \times 10^{- 4}$ g $/ 100$ mL of solution then the value of $K_{s p}$ will be: