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Solubility Product

The solubilit...

Question

The solubility product of NaCl is $2.5 \times 10^{- 9}$ $m o l e s^{2} / l i t^{2}$ . Its solubility is:

A

2.5 \times 10^{- 9} m o l / l i t

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B

5 \times 10^{- 5}

m o l e s^{2} / l i t^{2}

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C

5 \times 10^{- 5} m o l / l i t

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D

1.3 \times 10^{- 5} m o l / l i t

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Solution

The correct option is C $5 \times 10^{- 5} m o l / l i t$
$K_{s p} = s^{2}$
=> $s = \sqrt{K_{s p}} = 5 * 10^{- 5}$

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Similar questions

K_{s p} o f A g C l

= 2.8 \times 10^{- 10}

25^{\circ}

C. Calculate solubility of AgCl in 0.1 (M) NaCl

(K_{s p})

2.8 \times 10^{- 10}

25^{\circ}

C. Calculate solubility of AgCl in

0.1 (M) A g N O_{3}

Q. The solubility of aluminium sulphate is

1.0 \times 10^{- 4} m o l / l i t

, the solubility product is :

Q. The rate constant

(K)

for the reaction

2 A + B \to

product, was found to be

2.5 \times 10^{- 5}

l i t r e m o l^{- 1} s e c^{- 1}

15 s e c

2.60 \times 10^{- 5} l i t m o l^{- 1} s e c^{- 1}

30 s e c

2.55 \times 10^{- 5}

l i t m o l^{- 1} s e c^{- 1}

50 s e c

. The order of reaction is:

Q. For a particular

A + B \to C

25^{\circ} C

. The following results are obtained.

[A]	[B]	[C]
(mole/lit)	(moles/lit)	(mole lit $^{- 1} s e c^{- 2}$ )
$9 \times 10^{- 5}$	$1.5 \times 10^{- 2}$	$0.06$
$9 \times 10^{- 5}$	$3 \times 10^{- 3}$	$0.012$
$3 \times 10^{- 5}$	$3 \times 10^{- 3}$	$0.004$
$6 \times 10^{- 5}$	x	$0.024$

Then the value of x is :

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