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Question: Calculate the molar solubility of \(AgBr\;in\; 0.035MNaBr\left( {{K_{sp}} = 5 \times 1{0^{ - 13}}} \right)\).

Short Answer

Expert verified

The molar solubility is \(1.43 \cdot {10^{ - 11}}{\rm{M}}\) .

Step by step solution

01

Calculate the molar solubility:

The molar solubility is the chemical property of a solute and represents the maximum concentration of a solute in a solution at a certain temperature and pressure. Such a solution is called saturated.

AgBr dissociates according to the equation:

\(AgBr({\rm{s}}) \leftrightarrow {\rm{Ag}} + ({\rm{aq}}) + {\rm{Br}} - ({\rm{aq}})\)

The solubility product \({\rm{\;(Ksp) of\;AgBr}}\)

\({\rm{Ksp}} = ({\rm{Ag}} + )({\rm{Br}} – )\)

As we have the 0.035 M NaBr we can conclude that we have 0.035 M Br ions in the solution.

\(\begin{array}{*{20}{c}}{5.00 \times {{10}^{ - 13}} = ({\rm{Ag}} + ) \cdot 0.035{\rm{M}}}\\{({\rm{Ag}} + ) = \frac{{5.00 \times {{10}^{ - 13}}}}{{0.035{\rm{M}}}}}\end{array}\)

\(({\rm{Ag}} + ) = 1.43 \times {10^{ - 11}}{\rm{M}}\)

The solution for molar solubility is \(1.43 \cdot {10^{ - 11}}{\rm{M}}\)

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Most popular questions from this chapter

Magnesium hydroxide and magnesium citrate function as mild laxatives when they reach the small intestine. Why do magnesium hydroxide and magnesium citrate, two very different substances, have the same effect in your small intestine. (Hint: The contents of the small intestine are basic.)

Question: Using Lewis structures, write balanced equations for the following reactions:

\({\rm{\;}}(a)\;HCl(g) + P{H_3}(g) \to \)

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Question: The simplest amino acid is glycine, H2NCH2CO2H. The common feature of amino acids is that they contain the functional groups: an amine group, –NH2, and a carboxylic acid group, –CO2H. An amino acid can function as either an acid or a base. For glycine, the acid strength of the carboxyl group is about the same as that of acetic acid, CH3CO2H, and the base strength of the amino group is slightly greater than that of ammonia, NH3.

(a) Write the Lewis structures of the ions that form when glycine is dissolved in 1 M HCl and in 1 M KOH.

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

Question: 30. Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for \({K_{sp}}\) values.)

(a) \(KCl{O_4}:\left( {{K^ + }} \right) = 0.01{M^ - }\left( {ClO_4^ - } \right) = 0.01M\)

(b) \({K_2}PtC{l_6}:\left( {{K^ + }} \right) = 0.01M,\left( {PtC{l_6}^{2 - }} \right) = 0.01M\) \(\)

(c) \(Pb{I_2}:\left( {P{b^{2 + }}} \right) = 0.003M,\left( {{I^ - }} \right) = 1.3 \times 1{0^{ - 3}}M\)

(d) \(A{g_2}\;S:\left( {A{g^ + }} \right) = 1 \times 1{0^{ - 10}}M,\left( {{S^{2 - }}} \right) = 1 \times 1{0^{ - 13}}M\)

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