Solubility Equilibria of Sparingly Soluble Salts
CHXI07:EQUILIBRIUM

314955 Which expression correctly represents solubility product for,
\({\text{Ba(OH}}{{\text{)}}_{\text{2}}}{\text{.8}}{{\text{H}}_{\text{2}}}{\text{O(s)}} \rightleftharpoons {\text{B}}{{\text{a}}^{{\text{2 + }}}}{\text{(aq) + 2O}}{{\text{H}}^{{\text{ - }}}}{\text{(aq) + 8}}{{\text{H}}_{\text{2}}}{\text{O(l)?}}\)

1 \({{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ - }}}} \right]^{\rm{2}}}\)
2 \({{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{2O}}{{\rm{H}}^{\rm{ - }}}} \right]^{\rm{2}}}\)
3 \({{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ - }}}} \right]^{\rm{2}}}{\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]^{\rm{8}}}\)
4 \({{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ - }}}} \right]^{\rm{2}}}{\left[ {{\rm{8}}{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]^{\rm{8}}}\)
CHXI07:EQUILIBRIUM

314956 The solubility product expression for \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is represented as

1 \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^{3}\left[\mathrm{PO}_{4}^{3-}\right]^{2}\)
2 \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^{2}\left[\mathrm{PO}_{4}^{3-}\right]^{2}\)
3 \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{PO}_{4}^{2-}\right]^{3}\)
4 \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{PO}_{4}^{2-}\right]\)
CHXI07:EQUILIBRIUM

314957 The solubility product of a salt AB is \(1 \times 10^{-8}\). In a solution in which concentration of A is \(10^{-3} \mathrm{M}, \mathrm{AB}\) will precipitate when the concentration of B will be

1 \(10^{-7} \mathrm{M}\)
2 \(10^{-4} \mathrm{M}\)
3 \(10^{-5} \mathrm{M}\)
4 \(10^{-6} \mathrm{M}\)
CHXI07:EQUILIBRIUM

314958 \(\mathrm{M}(\mathrm{OH})_{\mathrm{x}}\) has \(\mathrm{K}_{\mathrm{sp}}=4 \times 10^{-12}\) and solubility \(10^{-4} \mathrm{M}\). Hence, \(\mathrm{x}\) is

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CHXI07:EQUILIBRIUM

314955 Which expression correctly represents solubility product for,
\({\text{Ba(OH}}{{\text{)}}_{\text{2}}}{\text{.8}}{{\text{H}}_{\text{2}}}{\text{O(s)}} \rightleftharpoons {\text{B}}{{\text{a}}^{{\text{2 + }}}}{\text{(aq) + 2O}}{{\text{H}}^{{\text{ - }}}}{\text{(aq) + 8}}{{\text{H}}_{\text{2}}}{\text{O(l)?}}\)

1 \({{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ - }}}} \right]^{\rm{2}}}\)
2 \({{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{2O}}{{\rm{H}}^{\rm{ - }}}} \right]^{\rm{2}}}\)
3 \({{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ - }}}} \right]^{\rm{2}}}{\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]^{\rm{8}}}\)
4 \({{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ - }}}} \right]^{\rm{2}}}{\left[ {{\rm{8}}{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]^{\rm{8}}}\)
CHXI07:EQUILIBRIUM

314956 The solubility product expression for \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is represented as

1 \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^{3}\left[\mathrm{PO}_{4}^{3-}\right]^{2}\)
2 \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^{2}\left[\mathrm{PO}_{4}^{3-}\right]^{2}\)
3 \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{PO}_{4}^{2-}\right]^{3}\)
4 \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{PO}_{4}^{2-}\right]\)
CHXI07:EQUILIBRIUM

314957 The solubility product of a salt AB is \(1 \times 10^{-8}\). In a solution in which concentration of A is \(10^{-3} \mathrm{M}, \mathrm{AB}\) will precipitate when the concentration of B will be

1 \(10^{-7} \mathrm{M}\)
2 \(10^{-4} \mathrm{M}\)
3 \(10^{-5} \mathrm{M}\)
4 \(10^{-6} \mathrm{M}\)
CHXI07:EQUILIBRIUM

314958 \(\mathrm{M}(\mathrm{OH})_{\mathrm{x}}\) has \(\mathrm{K}_{\mathrm{sp}}=4 \times 10^{-12}\) and solubility \(10^{-4} \mathrm{M}\). Hence, \(\mathrm{x}\) is

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2 2
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4 4
CHXI07:EQUILIBRIUM

314955 Which expression correctly represents solubility product for,
\({\text{Ba(OH}}{{\text{)}}_{\text{2}}}{\text{.8}}{{\text{H}}_{\text{2}}}{\text{O(s)}} \rightleftharpoons {\text{B}}{{\text{a}}^{{\text{2 + }}}}{\text{(aq) + 2O}}{{\text{H}}^{{\text{ - }}}}{\text{(aq) + 8}}{{\text{H}}_{\text{2}}}{\text{O(l)?}}\)

1 \({{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ - }}}} \right]^{\rm{2}}}\)
2 \({{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{2O}}{{\rm{H}}^{\rm{ - }}}} \right]^{\rm{2}}}\)
3 \({{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ - }}}} \right]^{\rm{2}}}{\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]^{\rm{8}}}\)
4 \({{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ - }}}} \right]^{\rm{2}}}{\left[ {{\rm{8}}{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]^{\rm{8}}}\)
CHXI07:EQUILIBRIUM

314956 The solubility product expression for \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is represented as

1 \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^{3}\left[\mathrm{PO}_{4}^{3-}\right]^{2}\)
2 \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^{2}\left[\mathrm{PO}_{4}^{3-}\right]^{2}\)
3 \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{PO}_{4}^{2-}\right]^{3}\)
4 \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{PO}_{4}^{2-}\right]\)
CHXI07:EQUILIBRIUM

314957 The solubility product of a salt AB is \(1 \times 10^{-8}\). In a solution in which concentration of A is \(10^{-3} \mathrm{M}, \mathrm{AB}\) will precipitate when the concentration of B will be

1 \(10^{-7} \mathrm{M}\)
2 \(10^{-4} \mathrm{M}\)
3 \(10^{-5} \mathrm{M}\)
4 \(10^{-6} \mathrm{M}\)
CHXI07:EQUILIBRIUM

314958 \(\mathrm{M}(\mathrm{OH})_{\mathrm{x}}\) has \(\mathrm{K}_{\mathrm{sp}}=4 \times 10^{-12}\) and solubility \(10^{-4} \mathrm{M}\). Hence, \(\mathrm{x}\) is

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NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
CHXI07:EQUILIBRIUM

314955 Which expression correctly represents solubility product for,
\({\text{Ba(OH}}{{\text{)}}_{\text{2}}}{\text{.8}}{{\text{H}}_{\text{2}}}{\text{O(s)}} \rightleftharpoons {\text{B}}{{\text{a}}^{{\text{2 + }}}}{\text{(aq) + 2O}}{{\text{H}}^{{\text{ - }}}}{\text{(aq) + 8}}{{\text{H}}_{\text{2}}}{\text{O(l)?}}\)

1 \({{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ - }}}} \right]^{\rm{2}}}\)
2 \({{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{2O}}{{\rm{H}}^{\rm{ - }}}} \right]^{\rm{2}}}\)
3 \({{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ - }}}} \right]^{\rm{2}}}{\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]^{\rm{8}}}\)
4 \({{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ - }}}} \right]^{\rm{2}}}{\left[ {{\rm{8}}{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]^{\rm{8}}}\)
CHXI07:EQUILIBRIUM

314956 The solubility product expression for \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is represented as

1 \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^{3}\left[\mathrm{PO}_{4}^{3-}\right]^{2}\)
2 \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^{2}\left[\mathrm{PO}_{4}^{3-}\right]^{2}\)
3 \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{PO}_{4}^{2-}\right]^{3}\)
4 \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{PO}_{4}^{2-}\right]\)
CHXI07:EQUILIBRIUM

314957 The solubility product of a salt AB is \(1 \times 10^{-8}\). In a solution in which concentration of A is \(10^{-3} \mathrm{M}, \mathrm{AB}\) will precipitate when the concentration of B will be

1 \(10^{-7} \mathrm{M}\)
2 \(10^{-4} \mathrm{M}\)
3 \(10^{-5} \mathrm{M}\)
4 \(10^{-6} \mathrm{M}\)
CHXI07:EQUILIBRIUM

314958 \(\mathrm{M}(\mathrm{OH})_{\mathrm{x}}\) has \(\mathrm{K}_{\mathrm{sp}}=4 \times 10^{-12}\) and solubility \(10^{-4} \mathrm{M}\). Hence, \(\mathrm{x}\) is

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4 4