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Chapter 15: Q26 E (page 872)

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, \(PbS{O_4}\)

Short Answer

Expert verified

The solubility product constant represented as Ksp can be defined as state in which a solid and its respective ions in given solution are in equilibrium. Its value indicates the extent to which a compound can undergo dissociation in water.

Step by step solution

01

To calculate the concentration of the ions in a saturated solution of TlCl

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

  1. \(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}} = \left[ {{\rm{T}}{{\rm{l}}^{\rm{ + }}}} \right]\left[ {{\rm{C}}{{\rm{l}}^{\rm{ - }}}} \right] = {{\rm{x}}^2} = 1.7 \times {10^{ - 4}}\\{\rm{x}} = 0.01\;{\rm{M}}\\\left[ {{\rm{T}}{{\rm{l}}^{\rm{ + }}}} \right] = \left[ {{\rm{C}}{{\rm{l}}^{\rm{ - }}}} \right] = 0.01\;{\rm{M}}\end{array}\)
02

To calculate the concentration of the ions in a saturated solution of

\(Ba{F_2}\)

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

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

03

To calculate the concentration of the ions in a saturated solution of

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

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

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

04

To calculate the concentration of the ions in a saturated solution of

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

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

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

05

To calculate the concentration of the ions in a saturated solution of

\(PbS{O_4}\)

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

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Most popular questions from this chapter

Question: The calcium ions in human blood serum are necessary for coagulation (Figure 15.5). Potassium oxalate, \({K_2}{C_2}{O_4}\), is used as an anticoagulant when a blood sample is drawn for laboratory tests because it removes the calcium as a precipitate of\(Ca{C_2}{O_4} \times {H_2}O\). It is necessary to remove all but 1.0% of the \(C{a^{2 + }}\) in serum in order to prevent coagulation. If normal blood serum with a buffered pH of 7.40 contains 9.5 mg of \(C{a^{2 + }}\) per 100 mL of serum, what mass of \({K_2}{C_2}{O_4}\)is required to prevent the coagulation of a 10 mL blood sample that is 55% serum by volume? (All volumes are accurate to two significant figures. Note that the volume of serum in a 10-mL blood sample is 5.5 mL. Assume that the \({K_{sp}}\)value for \(Ca{C_2}{O_4}\)in serum is the same as in water.)

What is the effect on the amount of\(CaHP{O_4}\) that dissolves and the concentrations of \(C{a^{2 + }}\;and\;HPO_4^ - \)when each of the following are added to a mixture of solid \(CaHP{O_4}\)and water at equilibrium?

\(\;(a)\;CaC{l_2}\)

\(\;(b)\;HCl\)

\(\;(c)\;KCl{O_4}\)

\(\;(d)\;NaOH\)

\(\;(e)\;CaHP{O_4}\)

Question: Explain why addition of \(N{H_3}\;or\;HN{O_3}\) to a saturated solution of \(Cu{(OH)_2}\)in contact with solid \(Cu{(OH)_2}\) increases the solubility of the solid.

Calculate the volume of 1.50MCH3CO2H required to dissolve a precipitate composed of 350mg each of and CaCO3,SrCO3,BaCO3

Question: Hydrogen sulfide is bubbled into a solution that is 0.10 M in both \(P{b^{2 + }}\)and \(F{e^{2 + }}\)and 0.30 M in HCl. After the solution has come to equilibrium it is saturated with \({H_2}S\) ((\({H_2}S\)) = 0.10 M). What concentrations of \(P{b^{2 + }}\)and \(F{e^{2 + }}\)remain in the solution? For a saturated solution of \({H_2}S\)we can use the equilibrium:
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