Solubility Equilibria of Sparingly Soluble Salts
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NEET DIGITAL LIBRARY
CHXI07:EQUILIBRIUM

314967 The solubility of \({\mathrm{\mathrm{PbCl}_{2}}}\) in water is \({\mathrm{\mathrm{S} \mathrm{~mol} \mathrm{~L}{ }^{-1}}}\) and its solubility product is \({\mathrm{\mathrm{K}_{\mathrm{sp}}}}\). The relation between \({\mathrm{\mathrm{K}_{\mathrm{sp}}}}\) and S is represented as \({\mathrm{\mathrm{S}=\sqrt[3]{\dfrac{\mathrm{K}_{\mathrm{sp}}}{x}}}}\). The value of \({\mathrm{x}}\) is ____ .

1 125
2 27
3 1
4 4
CHXI07:EQUILIBRIUM

314968 mole of \(\mathrm{BaCO}_{3}\) are soluble in \(1 \mathrm{~L}\), then x is related to its solubility product by the expression

1 \({\rm{x}} = {{\rm{K}}_{{\rm{sp }}}}\)
2 \({{\rm{x}}^{\rm{2}}} = \sqrt {{{\rm{K}}_{{\rm{sp}}}}} \)
3 \({\rm{x}} = \sqrt {{{\rm{K}}_{{\rm{sp}}}}} \)
4 \({\rm{x}} = \frac{{{{\rm{K}}_{{\rm{sp}}}}}}{2}\)
CHXI07:EQUILIBRIUM

314973 The solubility of \(\mathrm{AgCl}\) in it's solution is \(1.25 \times 10^{-5} \mathrm{moldm}^{-3}\). What is the solubility product of \(\mathrm{AgCl}\) ?

1 \(1.56 \times 10^{-10}\)
2 \(3.50 \times 10^{-6}\)
3 \(1.10 \times 10^{-5}\)
4 \(2.53 \times 10^{-3}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM

314974 For the electrolyte of type \({{\text{A}}_2}{\text{B}}\) solubility product is \(\mathrm{K}_{\mathrm{sp}}\). Then its solubility is

1 \(\dfrac{{{{\rm{K}}_{{\rm{sp}}}}}}{{\rm{4}}}\)
2 \(\sqrt[{\rm{3}}]{{\dfrac{{{{\rm{K}}_{{\rm{SP}}}}}}{{\rm{4}}}}}\)
3 \(\sqrt[{\rm{3}}]{{{{\rm{K}}_{{\rm{sp}}}}}}\)
4 \(\dfrac{{\sqrt {{{\rm{K}}_{{\rm{sp}}}}} }}{{\rm{4}}}\)
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CHXI07:EQUILIBRIUM

314967 The solubility of \({\mathrm{\mathrm{PbCl}_{2}}}\) in water is \({\mathrm{\mathrm{S} \mathrm{~mol} \mathrm{~L}{ }^{-1}}}\) and its solubility product is \({\mathrm{\mathrm{K}_{\mathrm{sp}}}}\). The relation between \({\mathrm{\mathrm{K}_{\mathrm{sp}}}}\) and S is represented as \({\mathrm{\mathrm{S}=\sqrt[3]{\dfrac{\mathrm{K}_{\mathrm{sp}}}{x}}}}\). The value of \({\mathrm{x}}\) is ____ .

1 125
2 27
3 1
4 4
CHXI07:EQUILIBRIUM

314968 mole of \(\mathrm{BaCO}_{3}\) are soluble in \(1 \mathrm{~L}\), then x is related to its solubility product by the expression

1 \({\rm{x}} = {{\rm{K}}_{{\rm{sp }}}}\)
2 \({{\rm{x}}^{\rm{2}}} = \sqrt {{{\rm{K}}_{{\rm{sp}}}}} \)
3 \({\rm{x}} = \sqrt {{{\rm{K}}_{{\rm{sp}}}}} \)
4 \({\rm{x}} = \frac{{{{\rm{K}}_{{\rm{sp}}}}}}{2}\)
CHXI07:EQUILIBRIUM

314973 The solubility of \(\mathrm{AgCl}\) in it's solution is \(1.25 \times 10^{-5} \mathrm{moldm}^{-3}\). What is the solubility product of \(\mathrm{AgCl}\) ?

1 \(1.56 \times 10^{-10}\)
2 \(3.50 \times 10^{-6}\)
3 \(1.10 \times 10^{-5}\)
4 \(2.53 \times 10^{-3}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM

314974 For the electrolyte of type \({{\text{A}}_2}{\text{B}}\) solubility product is \(\mathrm{K}_{\mathrm{sp}}\). Then its solubility is

1 \(\dfrac{{{{\rm{K}}_{{\rm{sp}}}}}}{{\rm{4}}}\)
2 \(\sqrt[{\rm{3}}]{{\dfrac{{{{\rm{K}}_{{\rm{SP}}}}}}{{\rm{4}}}}}\)
3 \(\sqrt[{\rm{3}}]{{{{\rm{K}}_{{\rm{sp}}}}}}\)
4 \(\dfrac{{\sqrt {{{\rm{K}}_{{\rm{sp}}}}} }}{{\rm{4}}}\)
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NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
CHXI07:EQUILIBRIUM

314967 The solubility of \({\mathrm{\mathrm{PbCl}_{2}}}\) in water is \({\mathrm{\mathrm{S} \mathrm{~mol} \mathrm{~L}{ }^{-1}}}\) and its solubility product is \({\mathrm{\mathrm{K}_{\mathrm{sp}}}}\). The relation between \({\mathrm{\mathrm{K}_{\mathrm{sp}}}}\) and S is represented as \({\mathrm{\mathrm{S}=\sqrt[3]{\dfrac{\mathrm{K}_{\mathrm{sp}}}{x}}}}\). The value of \({\mathrm{x}}\) is ____ .

1 125
2 27
3 1
4 4
CHXI07:EQUILIBRIUM

314968 mole of \(\mathrm{BaCO}_{3}\) are soluble in \(1 \mathrm{~L}\), then x is related to its solubility product by the expression

1 \({\rm{x}} = {{\rm{K}}_{{\rm{sp }}}}\)
2 \({{\rm{x}}^{\rm{2}}} = \sqrt {{{\rm{K}}_{{\rm{sp}}}}} \)
3 \({\rm{x}} = \sqrt {{{\rm{K}}_{{\rm{sp}}}}} \)
4 \({\rm{x}} = \frac{{{{\rm{K}}_{{\rm{sp}}}}}}{2}\)
CHXI07:EQUILIBRIUM

314973 The solubility of \(\mathrm{AgCl}\) in it's solution is \(1.25 \times 10^{-5} \mathrm{moldm}^{-3}\). What is the solubility product of \(\mathrm{AgCl}\) ?

1 \(1.56 \times 10^{-10}\)
2 \(3.50 \times 10^{-6}\)
3 \(1.10 \times 10^{-5}\)
4 \(2.53 \times 10^{-3}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM

314974 For the electrolyte of type \({{\text{A}}_2}{\text{B}}\) solubility product is \(\mathrm{K}_{\mathrm{sp}}\). Then its solubility is

1 \(\dfrac{{{{\rm{K}}_{{\rm{sp}}}}}}{{\rm{4}}}\)
2 \(\sqrt[{\rm{3}}]{{\dfrac{{{{\rm{K}}_{{\rm{SP}}}}}}{{\rm{4}}}}}\)
3 \(\sqrt[{\rm{3}}]{{{{\rm{K}}_{{\rm{sp}}}}}}\)
4 \(\dfrac{{\sqrt {{{\rm{K}}_{{\rm{sp}}}}} }}{{\rm{4}}}\)
NEET DIGITAL LIBRARY
CHXI07:EQUILIBRIUM

314967 The solubility of \({\mathrm{\mathrm{PbCl}_{2}}}\) in water is \({\mathrm{\mathrm{S} \mathrm{~mol} \mathrm{~L}{ }^{-1}}}\) and its solubility product is \({\mathrm{\mathrm{K}_{\mathrm{sp}}}}\). The relation between \({\mathrm{\mathrm{K}_{\mathrm{sp}}}}\) and S is represented as \({\mathrm{\mathrm{S}=\sqrt[3]{\dfrac{\mathrm{K}_{\mathrm{sp}}}{x}}}}\). The value of \({\mathrm{x}}\) is ____ .

1 125
2 27
3 1
4 4
CHXI07:EQUILIBRIUM

314968 mole of \(\mathrm{BaCO}_{3}\) are soluble in \(1 \mathrm{~L}\), then x is related to its solubility product by the expression

1 \({\rm{x}} = {{\rm{K}}_{{\rm{sp }}}}\)
2 \({{\rm{x}}^{\rm{2}}} = \sqrt {{{\rm{K}}_{{\rm{sp}}}}} \)
3 \({\rm{x}} = \sqrt {{{\rm{K}}_{{\rm{sp}}}}} \)
4 \({\rm{x}} = \frac{{{{\rm{K}}_{{\rm{sp}}}}}}{2}\)
CHXI07:EQUILIBRIUM

314973 The solubility of \(\mathrm{AgCl}\) in it's solution is \(1.25 \times 10^{-5} \mathrm{moldm}^{-3}\). What is the solubility product of \(\mathrm{AgCl}\) ?

1 \(1.56 \times 10^{-10}\)
2 \(3.50 \times 10^{-6}\)
3 \(1.10 \times 10^{-5}\)
4 \(2.53 \times 10^{-3}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM

314974 For the electrolyte of type \({{\text{A}}_2}{\text{B}}\) solubility product is \(\mathrm{K}_{\mathrm{sp}}\). Then its solubility is

1 \(\dfrac{{{{\rm{K}}_{{\rm{sp}}}}}}{{\rm{4}}}\)
2 \(\sqrt[{\rm{3}}]{{\dfrac{{{{\rm{K}}_{{\rm{SP}}}}}}{{\rm{4}}}}}\)
3 \(\sqrt[{\rm{3}}]{{{{\rm{K}}_{{\rm{sp}}}}}}\)
4 \(\dfrac{{\sqrt {{{\rm{K}}_{{\rm{sp}}}}} }}{{\rm{4}}}\)
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