Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Appendix J for solubility products).

(a) TlCl

(b) \(Ba{F_2}\)

(c) \(A{g_2}Cr{O_4}\)

(d) \(Ca{C_2}{O_4} \times {H_2}O\)

(e) The mineral anglesite, \(PbS{O_4}\)

a)K_sp of a saturated solution of TICl can be calculated as follows

\(Ba{F_2}\)

b) K_sp of a saturated solution of BaF₂ can be calculated as follows

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}} = \left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{{\rm{F}}^{\rm{ - }}}} \right]^2} = {\rm{x}}{(2{\rm{x}})^2} = 4{{\rm{x}}^3} = 2.4 \times {10^{ - 5}}\\{\rm{x}} = \sqrt[3]{{\frac{{2.4 \times {{10}^{ - 5}}}}{4}}} = 0.02\;{\rm{M}}\\\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right] = 0.02\;{\rm{M}}\\\left[ {{{\rm{F}}^{\rm{ - }}}} \right] = 0.04\;{\rm{M}}\end{array}\)

\(A{g_2}Cr{O_4}\)

C) K_sp of a saturated solution of AgCrO₄ can be calculated as follows

c) \(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}} = {\left[ {{\rm{A}}{{\rm{g}}^ + }} \right]^2}\left[ {{\rm{CrO}}_4^{2 - }} \right] = {(2{\rm{x}})^2}{\rm{x}} = 4{{\rm{x}}^3} = 9 \times {10^{ - 12}}\\{\rm{x}} = \sqrt[3]{{\frac{{9 \times {{10}^{ - 12}}}}{4}}} = 1.3 \times {10^{ - 4}}{\rm{M}}\\\left[ {{\rm{A}}{{\rm{g}}^ + }} \right] = 2.6 \times {10^{ - 4}}\\\left[ {{\rm{CrOH}}_4^{2 - }} \right] = 1.3 \times {10^{ - 4}}{\rm{M}}\end{array}\)

\(Ca{C_2}{O_4} \times {H_2}O\)

K_sp of\(Ca{C_2}{O_4} \times {H_2}O\) can be calculated as follows

d) \(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}} = \left[ {{\rm{C}}{{\rm{a}}^{2 + }}} \right]\left[ {{{\rm{C}}_2}{\rm{O}}_4^{2 - }} \right] \cdot \left[ {{{\rm{H}}_2}{\rm{O}}} \right] = {{\rm{x}}^3} = 1.96 \times {10^{ - 8}}\\{\rm{x}} = \sqrt[3]{{1.96 \times {{10}^{ - 8}}}}\;{\rm{M}} = 2.7 \times {10^{ - 3}}\;{\rm{M}}\\\left[ {{\rm{C}}{{\rm{a}}^{2 + }}} \right] = \left[ {{{\rm{C}}_2}{\rm{O}}_4^{2 - }} \right] = 2.7 \times {10^{ - 3}}\;{\rm{M}}\end{array}\)

\(PbS{O_4}\)

e) K_sp of a saturated solution of PbSO₄can be calculated as follows

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}} = \left[ {{\rm{P}}{{\rm{b}}^{{\rm{2 + }}}}} \right]\left[ {{\rm{SO}}_{\rm{4}}^{{\rm{2 - }}}} \right] = {{\rm{x}}^2} = 1.3 \times {10^{ - 8}}\\{\rm{x}} = 1.1 \times {10^{ - 4}}\;{\rm{M}}\\\left[ {{\rm{P}}{{\rm{b}}^{{\rm{2 + }}}}} \right] = \left[ {{\rm{SO}}_{\rm{4}}^{{\rm{2 - }}}} \right] = 1.1 \times {10^{ - 4}}\;{\rm{M}}\end{array}\)

Finally we get,

a) \(\left[ {{\rm{T}}{{\rm{l}}^ + }} \right] = \left[ {{\rm{C}}{{\rm{l}}^ - }} \right] = 0.01\;{\rm{M}}\)

b) \(\left[ {{\rm{B}}{{\rm{a}}^{2 + }}} \right] = 0.02\;{\rm{M}};\left[ {{{\rm{F}}^ - }} \right] = 0.04\;{\rm{M}}\)

c) \(\left[ {{\rm{A}}{{\rm{g}}^ + }} \right] = 2.6 \times {10^{ - 4}};\left[ {{\rm{Cr}}O_4^{2 - }} \right] = 1.3 \times {10^{ - 4}}\;{\rm{M}}\)

d) \(\left[ {{\rm{C}}{{\rm{a}}^{2 + }}} \right] = \left[ {{{\rm{C}}_2}{\rm{O}}_4^{2 - }} \right] = 2.7 \times {10^{ - 3}}\;{\rm{M}}\)

e) \(\left[ {{\rm{P}}{{\rm{b}}^{2 + }}} \right] = \left[ {{\rm{SO}}_4^{2 - }} \right] = 1.1 \times {10^{ - 4}}\;{\rm{M}}\)