Solubility Equilibria of Sparingly Soluble Salts
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CHXI07:EQUILIBRIUM

314979 Solubility \((s)\) of \(\mathrm{CaF}_{2}\) in terms of its solubility product is given as

1 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp }}}}} \right)^{1/3}}\)
2 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp}}}}/2} \right)^{1/3}}\)
3 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp}}}}/4} \right)^{1/3}}\)
4 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp}}}}/2} \right)^{1/2}}\)
CHXI07:EQUILIBRIUM

314980 The solubility product of a sparingly soluble salt \(\mathrm{AX}_{2}\) is \(3.2 \times 10^{-8}\). What is it's solubility in \(\mathrm{moldm}^{-3}\) ?

1 \(2.8 \times 10^{-4}\)
2 \(1.6 \times 10^{-5}\)
3 \(2.0 \times 10^{-3}\)
4 \(4.0 \times 10^{-4}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM

314981 Given the solubility product of \({{\text{A}}_{\text{3}}}{{\text{B}}_{\text{2}}}\) is \(2 \times 10^{-30}\). What will will be the solubility in moles/litre?

1 \(\left(1.85 \times 10^{-32}\right)^{1 / 5}\)
2 \(\left(\dfrac{2 \times 10^{-30}}{108}\right)^{1 / 5}\)
3 \(\left(\dfrac{10-28}{5400}\right)^{1 / 5}\)
4 All
CHXI07:EQUILIBRIUM

314982 If the solubility of calcium fluoride in pure water is \({\rm{x}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\rm{/}}{\mkern 1mu} {\rm{L}}\), its solubility product is

1 \(\sqrt {\text{2}} {\text{x}}\)
2 \({\text{2}}{{\text{x}}^{\text{2}}}\)
3 \({\text{4}}{{\text{x}}^{\text{3}}}\)
4 \({{\text{x}}^{\text{2}}}\)
CHXI07:EQUILIBRIUM

314983 The solubility of \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) is \(2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\) at \(298 \mathrm{~K}\). What is it's solubility product?

1 \(1.6 \times 10^{-6}\)
2 \(3.2 \times 10^{-11}\)
3 \(1.6 \times 10^{-11}\)
4 \(3.2 \times 10^{-6}\)
MHTCET - 2021
NEET DIGITAL LIBRARY
CHXI07:EQUILIBRIUM

314979 Solubility \((s)\) of \(\mathrm{CaF}_{2}\) in terms of its solubility product is given as

1 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp }}}}} \right)^{1/3}}\)
2 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp}}}}/2} \right)^{1/3}}\)
3 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp}}}}/4} \right)^{1/3}}\)
4 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp}}}}/2} \right)^{1/2}}\)
CHXI07:EQUILIBRIUM

314980 The solubility product of a sparingly soluble salt \(\mathrm{AX}_{2}\) is \(3.2 \times 10^{-8}\). What is it's solubility in \(\mathrm{moldm}^{-3}\) ?

1 \(2.8 \times 10^{-4}\)
2 \(1.6 \times 10^{-5}\)
3 \(2.0 \times 10^{-3}\)
4 \(4.0 \times 10^{-4}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM

314981 Given the solubility product of \({{\text{A}}_{\text{3}}}{{\text{B}}_{\text{2}}}\) is \(2 \times 10^{-30}\). What will will be the solubility in moles/litre?

1 \(\left(1.85 \times 10^{-32}\right)^{1 / 5}\)
2 \(\left(\dfrac{2 \times 10^{-30}}{108}\right)^{1 / 5}\)
3 \(\left(\dfrac{10-28}{5400}\right)^{1 / 5}\)
4 All
CHXI07:EQUILIBRIUM

314982 If the solubility of calcium fluoride in pure water is \({\rm{x}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\rm{/}}{\mkern 1mu} {\rm{L}}\), its solubility product is

1 \(\sqrt {\text{2}} {\text{x}}\)
2 \({\text{2}}{{\text{x}}^{\text{2}}}\)
3 \({\text{4}}{{\text{x}}^{\text{3}}}\)
4 \({{\text{x}}^{\text{2}}}\)
CHXI07:EQUILIBRIUM

314983 The solubility of \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) is \(2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\) at \(298 \mathrm{~K}\). What is it's solubility product?

1 \(1.6 \times 10^{-6}\)
2 \(3.2 \times 10^{-11}\)
3 \(1.6 \times 10^{-11}\)
4 \(3.2 \times 10^{-6}\)
MHTCET - 2021
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CHXI07:EQUILIBRIUM

314979 Solubility \((s)\) of \(\mathrm{CaF}_{2}\) in terms of its solubility product is given as

1 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp }}}}} \right)^{1/3}}\)
2 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp}}}}/2} \right)^{1/3}}\)
3 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp}}}}/4} \right)^{1/3}}\)
4 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp}}}}/2} \right)^{1/2}}\)
CHXI07:EQUILIBRIUM

314980 The solubility product of a sparingly soluble salt \(\mathrm{AX}_{2}\) is \(3.2 \times 10^{-8}\). What is it's solubility in \(\mathrm{moldm}^{-3}\) ?

1 \(2.8 \times 10^{-4}\)
2 \(1.6 \times 10^{-5}\)
3 \(2.0 \times 10^{-3}\)
4 \(4.0 \times 10^{-4}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM

314981 Given the solubility product of \({{\text{A}}_{\text{3}}}{{\text{B}}_{\text{2}}}\) is \(2 \times 10^{-30}\). What will will be the solubility in moles/litre?

1 \(\left(1.85 \times 10^{-32}\right)^{1 / 5}\)
2 \(\left(\dfrac{2 \times 10^{-30}}{108}\right)^{1 / 5}\)
3 \(\left(\dfrac{10-28}{5400}\right)^{1 / 5}\)
4 All
CHXI07:EQUILIBRIUM

314982 If the solubility of calcium fluoride in pure water is \({\rm{x}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\rm{/}}{\mkern 1mu} {\rm{L}}\), its solubility product is

1 \(\sqrt {\text{2}} {\text{x}}\)
2 \({\text{2}}{{\text{x}}^{\text{2}}}\)
3 \({\text{4}}{{\text{x}}^{\text{3}}}\)
4 \({{\text{x}}^{\text{2}}}\)
CHXI07:EQUILIBRIUM

314983 The solubility of \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) is \(2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\) at \(298 \mathrm{~K}\). What is it's solubility product?

1 \(1.6 \times 10^{-6}\)
2 \(3.2 \times 10^{-11}\)
3 \(1.6 \times 10^{-11}\)
4 \(3.2 \times 10^{-6}\)
MHTCET - 2021
For More Updates join with Us
CHXI07:EQUILIBRIUM

314979 Solubility \((s)\) of \(\mathrm{CaF}_{2}\) in terms of its solubility product is given as

1 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp }}}}} \right)^{1/3}}\)
2 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp}}}}/2} \right)^{1/3}}\)
3 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp}}}}/4} \right)^{1/3}}\)
4 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp}}}}/2} \right)^{1/2}}\)
CHXI07:EQUILIBRIUM

314980 The solubility product of a sparingly soluble salt \(\mathrm{AX}_{2}\) is \(3.2 \times 10^{-8}\). What is it's solubility in \(\mathrm{moldm}^{-3}\) ?

1 \(2.8 \times 10^{-4}\)
2 \(1.6 \times 10^{-5}\)
3 \(2.0 \times 10^{-3}\)
4 \(4.0 \times 10^{-4}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM

314981 Given the solubility product of \({{\text{A}}_{\text{3}}}{{\text{B}}_{\text{2}}}\) is \(2 \times 10^{-30}\). What will will be the solubility in moles/litre?

1 \(\left(1.85 \times 10^{-32}\right)^{1 / 5}\)
2 \(\left(\dfrac{2 \times 10^{-30}}{108}\right)^{1 / 5}\)
3 \(\left(\dfrac{10-28}{5400}\right)^{1 / 5}\)
4 All
CHXI07:EQUILIBRIUM

314982 If the solubility of calcium fluoride in pure water is \({\rm{x}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\rm{/}}{\mkern 1mu} {\rm{L}}\), its solubility product is

1 \(\sqrt {\text{2}} {\text{x}}\)
2 \({\text{2}}{{\text{x}}^{\text{2}}}\)
3 \({\text{4}}{{\text{x}}^{\text{3}}}\)
4 \({{\text{x}}^{\text{2}}}\)
CHXI07:EQUILIBRIUM

314983 The solubility of \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) is \(2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\) at \(298 \mathrm{~K}\). What is it's solubility product?

1 \(1.6 \times 10^{-6}\)
2 \(3.2 \times 10^{-11}\)
3 \(1.6 \times 10^{-11}\)
4 \(3.2 \times 10^{-6}\)
MHTCET - 2021
NEET DIGITAL LIBRARY
CHXI07:EQUILIBRIUM

314979 Solubility \((s)\) of \(\mathrm{CaF}_{2}\) in terms of its solubility product is given as

1 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp }}}}} \right)^{1/3}}\)
2 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp}}}}/2} \right)^{1/3}}\)
3 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp}}}}/4} \right)^{1/3}}\)
4 \({\rm{s}} = {\left( {{{\rm{K}}_{{\rm{sp}}}}/2} \right)^{1/2}}\)
CHXI07:EQUILIBRIUM

314980 The solubility product of a sparingly soluble salt \(\mathrm{AX}_{2}\) is \(3.2 \times 10^{-8}\). What is it's solubility in \(\mathrm{moldm}^{-3}\) ?

1 \(2.8 \times 10^{-4}\)
2 \(1.6 \times 10^{-5}\)
3 \(2.0 \times 10^{-3}\)
4 \(4.0 \times 10^{-4}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM

314981 Given the solubility product of \({{\text{A}}_{\text{3}}}{{\text{B}}_{\text{2}}}\) is \(2 \times 10^{-30}\). What will will be the solubility in moles/litre?

1 \(\left(1.85 \times 10^{-32}\right)^{1 / 5}\)
2 \(\left(\dfrac{2 \times 10^{-30}}{108}\right)^{1 / 5}\)
3 \(\left(\dfrac{10-28}{5400}\right)^{1 / 5}\)
4 All
CHXI07:EQUILIBRIUM

314982 If the solubility of calcium fluoride in pure water is \({\rm{x}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\rm{/}}{\mkern 1mu} {\rm{L}}\), its solubility product is

1 \(\sqrt {\text{2}} {\text{x}}\)
2 \({\text{2}}{{\text{x}}^{\text{2}}}\)
3 \({\text{4}}{{\text{x}}^{\text{3}}}\)
4 \({{\text{x}}^{\text{2}}}\)
CHXI07:EQUILIBRIUM

314983 The solubility of \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) is \(2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\) at \(298 \mathrm{~K}\). What is it's solubility product?

1 \(1.6 \times 10^{-6}\)
2 \(3.2 \times 10^{-11}\)
3 \(1.6 \times 10^{-11}\)
4 \(3.2 \times 10^{-6}\)
MHTCET - 2021