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Chapter 14: Q7 E (page 765)

Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: \(AgCl,BaS{O_4},Ca{F_2},H{g_2}{I_2},MnC{O_3},ZnS,PbS?\)

Short Answer

Expert verified

\(Ca{F_2},MnC{O_3},ZnS and PbS\) have a solubility bigger than that calculated from their solubility product.

Step by step solution

01

To find the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present

Of the following compounds:\(AgCl,BaS{O_4},Ca{F_2},H{g_2}{I_2},MnC{O_3},ZnS and PbS ,compounds Ca{F_2},MnC{O_3},ZnS and PbS\)will have a solubility bigger than that calculated from its solubility product.

02

Calculate solubility product due to hydrolysis of anion:

That is because the above slats are formed from anions of weak acids\(,{F^ - },CO_3^{ - 2},\;{S^{ - 2}}\) which get hydrated in water, that is, they trigger hydrolysis.

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Most popular questions from this chapter

Explain why a sample of pure water at \({40^ \circ }{\rm{C}}\) is neutral even though \(\left( {{{\rm{H}}_3}{{\rm{O}}^ + }} \right) = 1.7 \times {10^{ - 7}}M.\) \({K_{\rm{w}}}{\rm{\;is\;}}2.9 \times \)\({10^{ - 14}}{\rm{\;at\;}}{40^ \circ }{\rm{C}}.\)

For which of the following solutions must we consider the ionization of water when calculating the \(pH\) or \(pOH\)?

\((a) 3 \times 1{0^{ - 8}} M HN{O_3}\)

\((b) 0.10\;gHCl\)in \(1.0\;L\)of solution

\((c) 0.00080\;g NaOH\)in \(0.50\;L\)of solution

\((d) 1 \times 1{0^{ - 7}}M Ca{(OH)_2}\)

\((e) 0.0245 M KN{O_3}\)

From the equilibrium concentrations given, calculate for each of the weak acids and for each of the weak bases.

\(\begin{aligned}(a)C{H_3}C{O_2}H:\left( {{H_3}{O^ + }} \right) = 1.34 \times 1{0^{ - 3}}M;\left( {C{H_3}CO_2^ - } \right) = 1.34 \times 1{0^{ - 3}}M;\left( {C{H_3}C{O_2}H} \right) = 9.866 \times 1{0^{ - 2}}M;\\(b)Cl{O^ - }:\left( {O{H^ - }} \right) = 4.0 \times 1{0^{ - 4}}M;(HClO) = 2.38 \times 1{0^{ - 5}}M;\left( {Cl{O^ - }} \right) = 0.273M;\\(c)HC{O_2}H:\left( {HC{O_2}H} \right) = 0.524M;\left( {{H_3}{O^ + }} \right) = 9.8 \times 1{0^{ - 3}}M\left( {HCO_2^ - } \right) = 9.8 \times 1{0^{ - 3}}M;\\(d){C_6}{H_5}NH_3^ + :\left( {{C_6}{H_5}NH_3^ + } \right) = 0.233M;\left( {{C_6}{H_5}N{H_2}} \right) = 2.3 \times 1{0^{ - 3}}M;\left( {{H_3}{O^ + }} \right) = 2.3 \times 1{0^{ - 3}}M\end{aligned}\)

Calculate the \(pH\)and the\(pOH\) of each of the following solutions at \(2{5^o}C\) for which the substances ionize completely:

(a)\(0.000259M HCl{O_4}\)

(b)\(0.21M NaOH\)

(c)\(0.000071M Ba{(OH)_2}\)

(d) \(2.5M KOH\)

Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:

(a) \(HT{e^ - }\)(as a base)

(b) \({\left( {C{H_3}} \right)_3}N{H^ + }\)

(c) \(HAs{O_4}^{3 - }\)(as a base)

(d) \(HO_2^ - \)(as a base)

(e) \({C_6}{H_5}N{H_3}^ + \)

(f) \(HSO_3^ - \)(as a base)
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