The solubility product for a salt of type AB is $4×{10}^{-8}$.  The molarity of its standard solution will be:
1. $2×{10}^{-4}$ $\mathrm{mol}/\mathrm{L}$

2. $16×{10}^{-16}$ $\mathrm{mol}/\mathrm{L}$

3. $2×{10}^{-16}$ $\mathrm{mol}/\mathrm{L}$

4. $4×{10}^{-4}$ $\mathrm{mol}/\mathrm{L}$

Subtopic:  Solubility Product |
80%
From NCERT
NEET - 2020
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The solubility of $\mathrm{Ni}{\left(\mathrm{OH}\right)}_{2}$ in 0.1 M NaOH is:

Ksp ($\mathrm{Ni}{\left(\mathrm{OH}\right)}_{2}$)= 2x ${10}^{-15}$

1. 2 x ${10}^{-8}$ M.

2. 1 x ${10}^{-13}$ M

3. 1 x ${10}^{8}$

4. 2 x ${10}^{-13}$

Subtopic:  Solubility Product |
69%
From NCERT
NEET - 2020
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At room temperature, MY and NY3, two nearly insoluble salts, have the same Ksp values of 6.2 × 10-13. The true statement regarding MY and NY3 is:

 1 The molar solubility of MY in water is less than that of NY3. 2 The salts MY and NY3 are more soluble in 0.5 M KY than in pure water. 3 The addition of the salt of KY to a solution of MY and NY3 will have no effect on their solubilities. 4 The molar solubilities of MY and NY3 in water are identical.

Subtopic:  Solubility Product |
68%
From NCERT
NEET - 2016
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If the solubility of a ${\mathrm{M}}_{2}\mathrm{S}$ salt is $3.5$ $×$ ${10}^{-6}$ mol litre-1 , then the solubility product of ${\mathrm{M}}_{2}\mathrm{S}$ will be:

1.  $1.7$ $×$ ${10}^{-6}$ mollitre-3

2. $1.7$ $×$ ${10}^{-16}$ mol3 litre-3

3.  $1.7$ $×$ ${10}^{-18}$ mol3 litre-3

4. $1.7$ $×$ ${10}^{-12}$ mol3 litre-3

Subtopic:  Solubility Product |
72%
From NCERT
AIPMT - 2001
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The minimum volume of water required to dissolve 1g of calcium sulphate at 298 K is

(For CaSO4Ksp is 9.1 × 10–6)

1. 1.22 L

2. 0.69 L

3. 2.44 L

4. 1.87 L

Subtopic:  Solubility Product |
66%
From NCERT
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The molar solubility of ${\mathrm{CaF}}_{2}$ $\left({\mathrm{K}}_{\mathrm{sp}}=5.3$ $\mathrm{×}$ ${10}^{-11}\right)$ in 0.1 M solution of NaF will be:

 1 $5.3$ $\mathrm{×}$ ${10}^{-11}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$ 2 $5.3$ $\mathrm{×}$ ${10}^{-8}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$ 3 $5.3$ $\mathrm{×}$ ${10}^{-9}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$ 4 $5.3$ $\mathrm{×}$ ${10}^{-10}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$
Subtopic:  Solubility Product |
65%
From NCERT
NEET - 2019
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The solubility product of mercurous iodide is $4.5×{10}^{-29}$. The solubility of mercurous iodide will be:

1. $6.5×{10}^{-7}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

2. $4.09×{10}^{-8}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

3. $4.09×{10}^{-7}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

4. $6.5×{10}^{-8}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

Subtopic:  Solubility Product |
61%
From NCERT
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The solubility of BaSO4 in water is $2.42$ $×$ ${10}^{-3}$ g/ litre at 298 K. The value of the solubility product will be: (Molar mass of BaSO4 = 233 gmol–1)

 1 1.08 × 10–10 mol2 L–2 2 1.08 × 10–12 mol2 L–2 3 1.08 × 10–14 mol2 L–2 4 1.08 × 10–8 mol2 L–2
Subtopic:  Solubility Product |
61%
From NCERT
NEET - 2018
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A saturated solution of Ba(OH)2 has a pH of 12. The value of its Ksp will be:
1. $4.00×{10}^{-6}$ ${M}^{3}$
2. $4.00×{10}^{-7}$ ${M}^{3}$
3. $5.00×{10}^{-6}$ ${M}^{3}$
4. $5.00×{10}^{-7}$ ${M}^{3}$

Subtopic:  Solubility Product |
62%
From NCERT
AIPMT - 2010
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The maximum concentration of equimolar solutions, of ferrous sulphate and sodium sulphide, so that when mixed in equal volumes, there is no precipitation of iron sulphide, will be:

(For iron sulphide, Ksp = 6.3 × 10–18).

$1.$ $5.02$ $×{10}^{-9}$ $\mathrm{M}$

$2.$ $5.02$ $×$ ${10}^{9}$ $\mathrm{M}$

$3.$ $2.$ $25$ $×$ ${10}^{-13}$ $\mathrm{M}$

$4.$ $\mathrm{Can}\text{'}\mathrm{t}$ $\mathrm{predict}$

Subtopic:  Solubility Product |
56%
From NCERT
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