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Anomalous Solubility

Solubility pr...

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Solubility product of Agcl. Is 4.0 x 10 power-10 at 298 K .The solubility of Agcl in 0.04 M CaCl2 will be

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Q. The solubility product of

A g C l

4.0 \times 10^{- 10}

at 298 K. The solubility of

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{C a C l}_{2}

Q. Solubility product of

A g C l

1 \times 10^{- 6}

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. Its solubility in mole

l i t r e^{- 1}

Q. The solubility product of AgCl is

1.5625 \times 10^{- 10}

25^{0}

C. Its solubility in grams per litre will be :

Q. The solubility of

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298 K

1.06 \times 10^{- 5}

mole per litre. Calculate its solubility product at this temperature.

K_{s p}

2.8 \times 10^{- 10}

at a given temperature. The solubility of AgCl in 0.01 molar HCl solution at this temperature will be :

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