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Chapter 15: Q14 E (page 871)

Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:

\(\begin{array}{l}(a)A{g_2}S{O_4}\\(b)PbB{r_2}\\(c)AgI\\(d)Ca{C_2}{O_4} \times {H_2}O\end{array}\)

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.

a)\(x = 1.5 \times 1{0^{ - 2}}M\)

b) \(x = 2 \times 1{0^{ - 2}}M\)

c)\(x = 9.2 \times 1{0^{ - 9}}M\)

d) \(x = 6.3 \times 1{0^{ - 5}}M\)

Step by step solution

01

Step 1: To  calculate the molar solubility of each of the following from its solubility product:

The solubility product of\({\rm{A}}{{\rm{g}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{ = 1}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\)according to Ksp table

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}}\left( {{\rm{A}}{{\rm{g}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right){\rm{ = 1}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}{\left[ {{\rm{A}}{{\rm{g}}^{\rm{ + }}}} \right]^{\rm{2}}}\left[ {{\rm{SO}}_{\rm{4}}^{{\rm{2 - }}}} \right]\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\;{\rm{4}}{{\rm{x}}^{\rm{2}}}{\rm{ \times x}}\;\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\;{\rm{4}}{{\rm{x}}^{\rm{3}}}\\{\rm{x = }}\sqrt[{\rm{3}}]{{{{\rm{K}}_{{\rm{sp}}}}{\rm{ \div 4}}}}\\{\rm{x}}\;{\rm{ = }}\;\sqrt[{\rm{3}}]{{{\rm{1}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ \div 4}}}}\\{\rm{x}}\;{\rm{ = 1}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{M}}\end{array}\)

02

Step 2: To  calculate the molar solubility of each of the following from its solubility product:

The solubility product of\({\rm{PbB}}{{\rm{r}}_{\rm{2}}}{\rm{ = 4}}{\rm{.0 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\)according to Ksp table

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}}\left( {{\rm{PbB}}{{\rm{r}}_{\rm{2}}}} \right){\rm{ = 4}}{\rm{.0 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\left[ {{\rm{P}}{{\rm{b}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{B}}{{\rm{r}}^{\rm{ - }}}} \right]^{\rm{2}}}\\{\rm{ = }}\;{\rm{x \times 4}}{{\rm{x}}^{\rm{2}}}{\rm{ = 4}}{{\rm{x}}^{\rm{3}}}\\{\rm{x}}\;{\rm{ = }}\;\sqrt[{\rm{3}}]{{{{\rm{K}}_{{\rm{sp}}}}{\rm{ \div }}}}{\rm{4}}\\{\rm{x}}\;{\rm{ = }}\;\sqrt {{\rm{4}}{\rm{.0 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ \div 4}}} \\{\rm{x}}\;{\rm{ = }}\;{\rm{2 \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{M}}\end{array}\)

03

Step 3: To  calculate the molar solubility of each of the following from its solubility product:

The solubility product of\(\;(AgI) = 8.5 \times 1{0^{ - 17}}\)according to KSP table

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}}{\rm{(AgI) = 8}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 17}}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\left[ {{\rm{A}}{{\rm{g}}^{\rm{ + }}}} \right]\left[ {{{\rm{I}}^{\rm{ - }}}} \right]{\rm{ = }}\;{\rm{x \times x}}\;{\rm{ = }}\;{{\rm{x}}^{\rm{2}}}\\{\rm{x = }}\sqrt {{{\rm{K}}_{{\rm{sp}}}}} {\rm{ = }}\;\sqrt {{\rm{8}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 17}}}}} \\{\rm{x}}\;{\rm{ = }}\;{\rm{9}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{ - 9}}}}{\rm{M}}\end{array}\)

04

Step 4: To calculate the molar solubility of each of the following from its solubility product:

The solubility product of\(Ca{C_2}{O_4} = 4 \times 1{0^{ - 9}}\)according to KSP table

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}}\left( {{\rm{Ca}}{{\rm{C}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}} \right){\rm{ = 4 \times 1}}{{\rm{0}}^{{\rm{ - 9}}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\left[ {{\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}} \right]\left[ {{{\rm{C}}_{\rm{2}}}{\rm{O}}_{\rm{4}}^{{\rm{2 - }}}} \right]{\rm{ = x \times x = }}{{\rm{x}}^{\rm{2}}}\\{\rm{x}}\;{\rm{ = }}\sqrt {{{\rm{K}}_{{\rm{sp}}}}} {\rm{ = }}\sqrt {{\rm{4 \times 1}}{{\rm{0}}^{{\rm{ - 9}}}}} \\x = 6.3 \times 1{0^{ - 5}}M\end{array}\)

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Most popular questions from this chapter

What is the molar solubility of BaSO4 in a 0.250-M solution of NaHSO4? Ka for HSO4 − = 1.2×10–2.

What is the molar solubility of CaF2 in a 0.100 M solution of HF? Kafor HF = 7.2 × 10–4.

Question: In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO3 is added before precipitation begins. [The reaction of Ag+ with CN– goes to completion, producing the Ag(CN)2 − complex.] Precipitation of solid AgCN takes place when excess Ag+ is added to the solution, above the amount needed to complete the formation of Ag(CN)2 −. How many grams of NaCN were in the original sample?

Question: Under what circumstances, if any, does a sample of solid AgCl completely dissolve in pure water?

Question: 29. The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate \({K_{sp}}\) for each of the slightly soluble solids indicated:

(a) TlCl:\(\left( {T{l^ + }} \right) = 1.21 \times 1{0^{ - 2}}M,\left( {C{l^ - }} \right) = 1.2 \times 1{0^{ - 2}}M\)

(b)\(Ce{\left( {I{O_3}} \right)_4}:\left( {C{e^{4 + }}} \right) = 1.8 \times 1{0^{ - 4}}M,\left( {I{O_3}^ - } \right) = 2.6 \times 1{0^{ - 13}}M\)

(c)\(G{d_2}{\left( {S{O_4}} \right)_3}:\left( {G{d^{3 + }}} \right) = 0.132M,\left( {SO_4^{2 - }} \right) = 0.198M\)

(d)\(A{g_2}S{O_4}:\left( {A{g^ + }} \right) = 2.40 \times 1{0^{ - 2}}M,\left( {SO_4^{2 - }} \right) = 2.05 \times 1{0^{ - 2}}M\)

(e) \(BaS{O_4}:\left( {B{a^{2 + }}} \right) = 0.500M,\left( {SO_4^{2 - }} \right) = 2.16 \times 1{0^{ - 10}}M\)
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