91Ó°ÊÓ

\({K_{sp}} = 1.9 \times {10^{ - 4}} = \left( {T{l^ + }} \right)\left( {C{l^ - }} \right) = (x + 0.025) \times x\)

If we assume that\({\rm{x}}\)is small comparing to\(0.025\), then:

\(\begin{array}{l}{\rm{x}} = \frac{{1.9 \times {{10}^{ - 4}}}}{{0.025}}\\{\rm{x}} = 7.6 \times {10^{ - 3}}{\rm{M}}\end{array}\)

\({\rm{x}} = \frac{{7.6 \times {{10}^{ - 3}}}}{{0.025}} \times 10\% = 30\% \)

This change is significant, therefore we cannot drop the\({\rm{x}}\)value:

\({{\rm{x}}^2} + 0.025x - 1.9 \times {10^{ - 4}} = 0\)

\({\rm{x}} = \frac{{{\rm{ - b \pm }}\sqrt {{{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}}} }}{{{\rm{2a}}}} = \frac{{{\rm{ - 0}}{\rm{.025 \pm }}\sqrt {{\rm{6}}{\rm{.25 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ + 7}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}} }}{{\rm{2}}}{\rm{ = 0}}{\rm{.0061}}\;{\rm{M}}\)

\(\left( {{\rm{T}}{{\rm{l}}^{\rm{ + }}}} \right) = 0.025 + 0.0061 = 3.1 \times {10^{ - 2}}{\rm{M}}\)

\(\;\;\;\left( {{\rm{C}}{{\rm{l}}^ - }} \right) = 6.1 \times {10^{ - 3}}{\rm{M}}\)

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}} = 1.7 \times {10^{ - 6}}\\{{\rm{K}}_{{\rm{sp}}}} = \left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{{\rm{F}}^{\rm{ - }}}} \right]^2}\\{{\rm{K}}_{{\rm{sp}}}} = x \times {(x + 0.0313)^2}\end{array}\)

If we assume that x is small comparing to\(0.0313\), then

\(\begin{array}{l}{\rm{x}} = \frac{{1.7 \times {{10}^{ - 6}}}}{{{{0.0313}^2}}}\\{\rm{x}} = 1.7 \times {10^{ - 3}}\end{array}\)\(\frac{{1.7 \times {{10}^{ - 3}}}}{{0.0313}} \times 100\% = 5.5\% \)

This change is significant, therefore we cannot drop the\(x\)value:

By solving the equation, we get:

\(\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right] = 1.6 \times {10^{ - 3}}{\rm{M}}\)

\(\left[ {{{\rm{F}}^{\rm{ - }}}} \right] = 1.6 \times {10^{ - 3}} + 0.0313 = 0.0329\;{\rm{M}}\)

First, we need to find the molar concentration of the\({\rm{Mg}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\) :

\(\begin{array}{l}{\rm{M}} = \frac{{8.156\;{\rm{g}}}}{{148.3149\;{\rm{g}}/{\rm{mol}}}} = 0.05499\;{\rm{M}}\\{\rm{M}} = \frac{{0.05499\;{\rm{mol}}}}{{2.25\;{\rm{L}}}} = 0.02444{\rm{M}}\end{array}\)

If we assume that \({\rm{x}}\)is small comparing to\(0.02444\;{\rm{M}}\), then:

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}}{\rm{ = 8}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\left[ {{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}} \right]\left[ {{{\rm{C}}_{\rm{2}}}{\rm{O}}_{\rm{4}}^{{\rm{2 - }}}} \right]\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\;{\rm{x \times (x + 0}}{\rm{.02444)}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\;{\rm{0}}{\rm{.02444}} \times {\rm{x}}\end{array}\)\(\)

\(x = \left[ {{{\rm{C}}_2}{\rm{O}}_4^{2 - }} \right] = 3.5 \times {10^{ - 3}}\)

\(\begin{array}{l} = \frac{{3.5 \times {{10}^{ - 3}}}}{{0.02444}} \times 100\% \\ = 14\% \end{array}\)

This change is significant, therefore we cannot drop the\(x\)value.

\({{\rm{x}}^2} + 0.02444x - 8.6 \times {10^{ - 5}} = 0\)

\(\begin{array}{l}{\rm{x}} = \frac{{ - {\rm{b \pm }}\sqrt {{{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}}} }}{{2{\rm{a}}}}\\{\rm{x}} = \frac{{ - 0.02444 \pm \sqrt {5.973 \times {{10}^{ - 4}} + 3.44 \times {{10}^{ - 4}}} }}{2}\\{\rm{x}} = 3.5 \times {10^{ - 3}}{\rm{M}}\end{array}\)

\(\left[ {{{\rm{C}}_2}{\rm{O}}_4^{2 - }} \right] = 3.5 \times {10^{ - 3}}{\rm{M}}\)

\(\left[ {{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}} \right] = 3.1 \times {10^{ - 3}} + 0.02444 = 0.0275\;{\rm{M}}\)

\(\begin{array}{l}{\rm{pH}} = 12.7;\\{\rm{pOH}} = 1.3;[{\rm{O}}{{\rm{H}}^ - }] = 0.051\;{\rm{M}}\\{{\rm{K}}_{sp}} = 7.9 \times {10^{ - 6}} = \left[ {{\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ - }}}} \right]^2} = x \times {(x + 0.05)^2}\end{array}\)

If we assume that\({\rm{x}}\)is small comparing to\(0.05{\rm{M}}\), then:\(x = \frac{{7.9 \times {{10}^{ - 6}}}}{{{{0.05}^2}}} = 3.16 \times {10^{ - 3}}{\rm{M}}\)\(\frac{{3.16 \times {{10}^{ - 3}}}}{{0.05}} \times 100\% = 6.28\% \)This change is significant, therefore we cannot drop the\({\rm{x}}\)value.

By solving the equation, we get: \(\left[ {C{a^{2 + }}} \right] = 2.8 \times {10^{ - 3}}{\rm{M}}\)

\(\begin{array}{l}\left[ {{\rm{O}}{{\rm{H}}^{\rm{ - }}}} \right] = \left( {2.8 \times {{10}^{ - 3}} + 0.0501} \right)\\\left[ {{\rm{O}}{{\rm{H}}^{\rm{ - }}}} \right] = 0.053 \times {10^{ - 2}}{\rm{M}}\end{array}\)

Finally we get,

a) \(\left[ {{\rm{T}}{{\rm{l}}^ + }} \right] = 3.1 \times {10^{ - 2}}{\rm{M}};\left[ {{\rm{C}}{{\rm{l}}^ - }} \right] = 6.1 \times {10^{ - 3}}{\rm{M}}\)

b) \(\begin{array}{l}\left[ {{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}} \right] = 1.6 \times {10^{ - 3}}{\rm{M}};\\\left[ {{{\rm{F}}^{\rm{ - }}}} \right] = 0.0329{\rm{M}}\end{array}\)

c) \(\begin{array}{l}\left[ {{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}} \right]{\rm{ = }}\;{\rm{0}}{\rm{.0275}}\;{\rm{M}}\\\left[ {{{\rm{C}}_{\rm{2}}}{\rm{O}}_{\rm{4}}^{{\rm{2 - }}}} \right]{\rm{ = }}\;{\rm{3}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{M}}\end{array}\)

d) \(\begin{array}{l}\left[ {{\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}} \right]\;{\rm{ = }}\;{\rm{2}}{\rm{.8 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{M}}\\\left[ {{\rm{O}}{{\rm{H}}^{\rm{ - }}}} \right]{\rm{ = }}\;{\rm{0}}{\rm{.053 \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{M}}\end{array}\)