Skip to main content
Chemistry LibreTexts

17: Solubility and Complex ion Equilibria

These are homework exercises to accompany the Textmap created for "General Chemistry: Principles and Modern Applications " by Petrucci et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.


Write \(K_{sp}\) expression for the following equilibria.

  1. \(PbI_2 (s) \rightleftharpoons Pb^{2+}(aq)+2I^-_{(aq)}\)
  2. \(Cu(IO_3)_{2\; ­(s)} \rightleftharpoons Cu^{2+}­(aq)+2IO^-_{3\;_(aq)}\)
  3. \(Ag(OH)_{3\; (s)} \rightleftharpoons Al^{3+}(aq)+3OH^-_{(aq)}\)
  4. \(CaF_{2\; (s)} \rightleftharpoons Ca^{2+}_{(aq)}+2F^-_{(aq)}\)


  1. \(K_{sp} = [Pb^{2+}][I^-]^2\)
  2. \(K_{sp}= [Cu^{2+}][IO_3^-]^2\)
  3. \(K_{sp}= [Al^{3+}][OH^-]^3\)
  4. \(K_{sp}= [Ca^{2+}][F^-]^2\)


Write \(K_{sp}\) expression for the following equilibria. For example \(MX_{2 (s)} \rightleftharpoons  M^{2+}_{ (aq)} + 2 X^-_{ (aq)}\), \(K_{sp}= [M^{2+}][X^-]^2\)

  • \(Al(OH)_{3 (s)} \rightleftharpoons  Al^{3+}_{ (aq)} + 3 OH^-_{ (aq)}\)
  • \(CaF_{2 (s)} \rightleftharpoons Ca^{2+}_{ (aq)} + 2F^-_{ (aq)}\)
  • \(AgI _{(s)} \rightleftharpoons Ag^+_{ (aq)} + I^-_{ (aq)}\)
  • \(BaSO_{4 (s)} \rightleftharpoons  Ba^{2+}_{ (aq)} + SO^{2-}_{4(aq)}\)


  1. \(K_{sp}= [Al^{3+}][OH^-]^3\)
  2. \(K_{sp}=[Ca^{2+}][F^-]^2\)
  3. \(K_{sp}=[Ag^+][I^-]\)
  4. \(K_{sp}=[Ba^{2+}][SO_4^{2+}]\)


Write \(K_{sp}\) expressions for the following equilibrium reactions

  1. \(AgBr_{(s)} \rightleftharpoons Ag^+_{(aq)}+ Br^-_{(aq)}\)
  2. \(Al(OH)_{3(s)} \rightleftharpoons Al^{3+}_{(aq)}+3OH^-_{(aq)}\)
  3. \(PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)}+2Cl^-_{(aq)}\)


  1. \(K_{sp}=[Ag^+ ][Br^-]\)
  2. \(K_{sp}=[Al^{3+} ][OH^-]^3\)
  3. \(K_{sp}=[Pb^{2+}][Cl^-]^2\)


What would the \(K_{sp}\) expression be for the following reactions?

  1. \(NH_4Br_{(s)} \rightleftharpoons  NH^+_{4(aq)} + Br^-_{(aq)}\)
  2. \(BaCrO_{4(s)} \rightleftharpoons  Ba^{2+}_{(aq)} + CrO^{2-}_{4(aq)}\)
  3. \(Na_2S_2O_{3(s)}\rightleftharpoons  2Na^+_{(aq)} + S_2O_{3(aq)}^{2-}\)
  4. \(Fe_2(Cr_2O_7)_{3(s)} \rightleftharpoons  2Fe^{3+}_{(aq)}+ 3Cr_2O_{7(aq)}^{2-}\)


  • \([NH_4^+][Br^-]\)
  • \([Ba^{2+}][CrO_4^{2-}]\)
  • \([Na^+]^2[S_2O_3^{2-}]\)
  • \([Fe^{3+}]^2[Cr_2O_7^{2-}]^3\)


What is the solubility equilibrium and solubility-product constant expression for \(CaF_2\)?


1. \(CaF_{2(s)} \rightleftharpoons Ca^{2+}_{(aq)} + 2F^-_{(aq)}\)

2. \(K_{sp} = [Ca^{2+}][F^-]^2\)


Write the solubility equilibrium equations that are described by the following \(K_{sp}\) expressions. For example, \(K_{sp}= [Ra^{2+}][IO_3^-]^2\) represents \(Ra(IO_ 3)_2 \rightleftharpoons Ra^{2+}_{(aq)} + 2IO^-_{3(aq)}\)

  1. \(PbCl_{2(s)} \rightleftharpoons  Pb^{2+}_{(aq)} + 2 Cl^-_{(aq)}\)
  2. \(Mg_3(PO_4)_{2(s)} \rightleftharpoons 3 Mg^{2+}_{(aq)} + 2 PO_{4(aq)}^{3-}\)    
  3. \(SrCO_{3(s)} \rightleftharpoons Sr^{2+}_{(aq)} + CO^{2-}_{(aq)}\)    
  4. \(Hg_2Cl_{2(s)} \rightleftharpoons Hg^{2+}_{(aq)} + 2 Cl^-_{(aq) }\)


  1. \(K_{sp}= [Pb^{2+}][Cl^-]^2\)
  2. \(K_{sp}= [Mg^{2+}]^3[PO_4^{3-}]^2\)
  3. \(K_{sp}= [Sr^{2+}][CO_3^{2-}]\)
  4. \(K_{sp}= [Hg_2^{2+}][Cl^-]^2\) 


Write the equilibrium reactions for these \(K_{sp}\) expressions

  1. \(K_{sp}=[Ag^+][CrO_4^{2-}]\)
  2. \(K_{sp}= [Ba^{2+} ][SO_4^{2-} ]\)
  3. \(K_{sp}=[Pb^{2+} ][CrO_4^{2-}]\)


  1. \(Ag_2 CrO_{4(s)} \rightleftharpoons Ag^+_{(aq)}+ CrO^{2-}_{4(aq)}\)
  2. \(BaSO_{4(s)} \rightleftharpoons Ba^{2+}_{(aq)}+ SO^{2-}_{4(aq)}\)
  3. \(PbCrO_{4(s)} \rightleftharpoons Pb^{2+}_{(aq)}+ CrO_{4(aq)}^{2-}\)


Write the solubility equilibrium equations according to the \(K_{sp}\) expression provided.

  1. \(AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)}+Cl^-_{(aq)}\)
  2. \(Ra(IO_3)_{2(s)} \rightleftharpoons Ra^{2+}_{(aq)}+2IO^-_{3(aq)}\)
  3. \(Ag_2CrO_{4(s)} \rightleftharpoons 2Ag^+_{(aq)} + CrO^{2-}_{4(aq)}\)
  4. \(MgF_{2(s)} \rightleftharpoons Mg^{2+}_{(aq)} + 2F^-_{(aq)}\)


  1. \(K_{sp}  = [Ag^+ ][Cl^-]\)
  2. \(K_{sp}  = [Ra^{2+}] [IO­_3^-]^2\)
  3. \(K_{sp}  = [Ag^+]^2[CrO_4^{2-}]\)
  4. \(K_{sp}  = [Mg^{2+}][F^-]^2\)


Write the solubility product expression to which each one applies:

  1. \(K_{sp}    (Hg_2Cl_2) = 1.3 \times 10^{-19}\)
  2. \(K_{sp}   (AgI) = 8.5 \times 10^{-17}\)
  3. \(K_{sp}  (SrSO_4) = 3.2 \times 10^{-7}\)
  4. \(K_{sp}   (Fe(OH)_3) = 4 \times 10^{-38}\)


  1. \(K_{sp}  = [Hg^{2+}][Cl^-]^2=1.3 \times 10^{-19}\)
  2. \(K_{sp}  = [Ag^+][I^-]= 8.5 \times 10^{-17}\) 
  3. \(K_{sp} = [Sr^{2+}][SO_4^{2-}]=3.2 \times 10^{-7}\)
  4. \(K_{sp}  = [Fe^{3+}][OH^-]^3 = 4 \times 10^{-38}\)


The following Ksp values are found in a table. Write the solubility product expression to which each one applies. For example, \(K_{sp} (BaCO_3) = [Ba^{2+}][CO_3^-]= 5.1\times 10^{-9}\)

  1. \(K_{sp}(CaSO_4) = 9.1\times 10^{-6}\)
  2. \(K_{sp}(PbI_2)=7.1\times 10^{-9}\)
  3. \(K_{sp}(Mg(OH)_2)=1.8\times 10^{-11}\)
  4. \(K_{sp}(Ag_2CrO_2)=1.1\times 10^{-12}\)


  1. \([Ca^{2+}][SO_4^{2-}]=9.1\times 10^{-6}\)
  2. \([Pb^{2+}][I^-]^2=7.1\times 10^{-9}\)
  3. \([Mg^{2+}][OH^-]^2=1.8\times 10^{-11}\)
  4. \([Ag^+]^2[CrO_4^{2-}]=1.1\times 10^{-12}\)


Write the solubility product expression for each \(K_{sp}\) value

  1. \(K_{sp} (BaF_2 )=3.15 \times 10^{-3}\)
  2. \(K_{sp} (Mg(OH)_2 )= 1.71 \times10^{-4}\)
  3. \(K_{sp} (BaCO_3 ) = 5.1 \times 10^{-9}\)


  1. \(K_{sp}=[Ba^+] [F^-]^2=3.15 \times 10^{-3}\)
  2. \(K_{sp}=[Mg^{2+}] [OH^- ]^2= 1.71 \times10^{-4}\)
  3. \(K_{sp}=[Ba^{2+}][CO_3^{2-}]= 5.1 \times 10^{-9}\)


Write the solubility product expression for the following.

  1. \(Ag_2CrO_{4(s)} \)
  2. \(Mg_3(PO_4)_{2 (s)}\)
  3. \(PbCl_{2(s)} \)
  4. \(Cr(OH)_3\)


  1. \([CrO_4^{2-}][Ag^+]^2\)
  2. \([Mg^{2+}]^3[PO_4^{3-}]^2\)
  3. \([Pb^{2+}][Cl^-]^2\)
  4. \([Cr^{3+}][OH^-]^3\)


Which of the following would be more soluble in an acidic solution: \(Zn(OH)_2\), \(KBr\), \(AlCl_3\), \(PbS\)?


\(Zn(OH)_2\) and \(PbS\) would be more soluble in an acidic solution.


A particular water sample that is saturated in \(CaF_2\) has a \(Ca^{2+}\) content of 130 ppm (i.e., 130 g \(Ca^{2+}\) per \(10^6\) grams of water sample). What is the \(F^-\) ion content of the water in ppm?


\(CaF_2\) dissociates into \(Ca^{2+}\) and \(2F^-\). \((CaF_2\rightleftharpoons Ca^{2+} + 2F^-)\) So the ion content of the \(F^-\) ion would be double that of the \(Ca^{2+}\) ion content per \(10^6\) grams of the water sample. \(\frac{260\ g\ F^-}{10^6\ g} = 260 ppm\)


How much \(SO_4^{2-}\) ion content (in ppm) that a sample of water contains when it’s saturated in \(CaSO_4\) and it has a \(Ca^{2+}\) content of 40 ppm?


\([SO_4^{2-}] = 2.76\; M = 2.65 \times 10^{-7}\; ppm\)

Video Solution


A sample of water is saturated with \(Mg(OH)_2\). The sample has a \(Mg^{2+}\) content of 45 ppm (that is, 45 g \(Mg^{2+}\) per 106 g of water sample. What is the \(OH^-\) ion content of the water in ppm?


\([Mg^{2+}]= \frac{45\ g\ Mg^{2+}}{10^{6}\ g\ soln}\times \frac{1\ mol\ Mg^{2+}}{24.31\ g\ Mg^{2+}}\times \frac{1000g\ soln}{1\ L\ soln} = 1.85\times 10^{-3}\)

\([Mg^{2+}][OH^-]^2 = K_{sp} = 1.8\times 10^{-11} = (1.85\times 10^{-3})[OH^-]^2\)

\([OH^-]= 9.86\times 10^{-5}\)

\(ppm\ OH^- = \frac{9.86\times 10^{-5}}{1\ L\ soln}\times \frac{17\ g\ OH^-}{1\ mol\ OH^-}\times \frac{1\ L\ soln}{1000\ g}\times 10^6\ soln = 1.68 ppm\)


Silver chromate is dissolved in 0.1M \(AgNO_3\). What affect does the common ion have on the solubility? \(K_{sp} = 9 \times 10^{-12}\) molar solubility =\(1.3\times 10^{-4}\ M\)


\[Ag_2CrO_{4(s)} \rightleftharpoons 2Ag^+_{(aq)} + CrO^{2-}_{4 (aq)}\]

ICE \(Ag_2CrO_{4(s)}^+\) \(2Ag^+_{(aq)}\) \(CrO^{2-}_{4 (aq)}\)
Initial - 0.1 M 0
Change - +2x +x
Equilibrium - 0.1M+2x x

\[K_{sp}=[Ag_+]^2[CrO_4^{2-}]=  (0.1\;M+2x)^2(x)= 9 \times 10^{-12}\]

if we assume \(x <<  (0.1M)^2(x)=9 \times 10^{-12} \approx 9 \times 10^{-10}\)


Will a precipitate of lead (II) occur if three drops of 0.20MKI are added to 100.0mL of 0.010M \(Pb(NO_3)_2\)? assuming 1 drop= 0.05 mL.


\(PbI_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)}+2I^-_{(aq)}\)

\(K_{sp}=7.1\times 10^{-9}\)

\(amtI-=(3drops)\times \frac{0.05mL}{1drop}\times \frac{L}{1000mL}\times \frac{0.20mol\ KI}{L}\times \frac{mol\ I^-}{mol\ KI}=3.0\times 10^{-5} mol\ I^-\)

\(total\ volume=(0.1L)+3drops\frac{0.05mL}{1drop} \frac{L}{1000mL}=0.10015L\)

\([I^-]=\frac{3.0\times 10^-5\ mol}{0.10015L}=3.0\times 10^{-4}\ M\)

\(Q_{sp}=[Pb^{2+}][I^-]^2=(0.010M)(3.0\times 10^{-4})^2=9.0\times 10^{-10}\)

Since \(Q_{sp} < K_{sp}\), therefore, \(PbI_{2(s)}\) will NOT precipitate


Can \(Ce(IO_3)_3\) precipitate when adding together 100 mL of \(1.0 \times 10^{-2} \; M\) \(Ce(NO_3)_3\) and 200 mL of \(2.0 \times 10^{-3} \;M\) \(KIO_3\)? (\(K_{sp} = 1.9 \times 10^{-10}\)).


Yes, it will precipitate, since Q is bigger than K.


4.7 mL of 0.35 M \(KI\) are added to a 150.0 mL of 0.069 M \(Pb(NO_3)_2\). Will a precipitate of lead iodide be formed?

The following reactions has a \(K_{sp}\) of \(7.1 \times 10^{-9}\):

$$PbI_{2 (s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2 I^-_{(aq)}$$


Two salt solutions are combined so four different salts may precipitate out. However, we are only interseted in the \(Pb(NO_3)_2\). So this is a simple compare problem involving calculating \(Q_{sp}\) and then comparing to \(K_{sp}\).

\[Q_{sp} =  [Pb^{2+}][I^-]^2 \]

Both \([Pb^{2+}]\) and \([I^-]\) are needed to be known to calculate \(Q_{sp}\) for the final solution. This is a changing concentration problem too.

\[(4.7\; \cancel{mL} ) \left(\dfrac{1\; L}{1000\; \cancel{mL}}\right) \left( \dfrac{ 0.35\; \cancel{mol \;KI}}{1\; L}\right) \left( \dfrac{1\; mol\; I^-}{1 \; \cancel{mol\; KI}}\right) = 1.65 \times 10^{-3} mol I^-\]

Convert to concentration

\[[I^-] = \dfrac{1.65 \times 10^{-3}\, mol\, I^-}{0.150\; L} = 1.07 \times 10^{-2}\; M\]

\[Q_{sp}= [Pb^{2+}][I^-]^2 = (0.069)(1.07 \times 10^{-2})^2=8.3 \times 10^{-6} > K_{sp}=7.1 \times 10^{-9}\]

Therefore, \(PbI_{2(s)}\) should precipitate after mixing these two solutions.


When 310 ml of 0.460 M \(CaS\) are added to 310.0 ml of 0.112M \(NaNO_3\), what percentage of the \(Na^+\) is left unprecipitated?


\[NaNO_3 + CaS  \rightarrow NaS + CaNO_3\]

\[(310\; mL)(0.460\;M \;CaS) = 142.6\; mmol \;CaS\]

\[(310\; mL)(0.112\; M\; NaNO_3)= 34.72\; mmol\; NaNO_3\]

There are no unprecipitated sodium ions in the solution because there is an excess of CaS.


How much \(Mg^{2+}\) ions will remain in solution after adding \(NaOH\) to a solution that’s composed of 200.0 mL of 0.010M \(Mg^{2+}\) and 0.005M \(Ca^{2+}\)?


The \(Mg^{2+}\) ions already get used 100%.


When 70.0 mL sample of 0.372 M \(Na_2SO_{4(aq)}\) is added to 100 mL of 0.138 M \(Ca(NO_3)_{2(aq)}\), what percentage of \(Ca^{2+}\) remains unprecipitated?


\([SO_4^{2-}]= 0.372\ M \times \frac{70\ mL}{(70mL+100mL)}=  0.153\ M\)

\([Ca^{2+}]= 0.138\ M \times \frac{100mL}{(70mL+100mL}= 0.0776 M\)

We determine the value of the ion product and compare it to the solubility product constant value.

\(Q_{sp}= [Ca^{2+}][NO_3^-]^2 = (0.0776)(0.153)^2= 1.82\times 10^{-3} > 9.1\times 10^{-6} = K_{sp}\ for\ Ca(NO_3)_2\)

Since \(Q_{sp} > K_{sp}\), \(Ca(NO_3)_2\) should precipitate.


Equation:             \(Ca(NO_3)_{2(s)}   \rightleftharpoons \quad  Ca^{2+}_{(aq)} + \quad 2NO_{3(aq)}^-\)

Orig. soln:              ____         0.0776 M                   0.153M

Form solid:            ____       -0.0776 M                 -0.00219M

Not @ Eq:              ____             0 M                        0.151 M

Changes:               ____         + X M                        + 2X M

Equil:                      ____             X M                   (0.151 + 2X) M


\(K_{sp}= [Ca^{2+}][NO_3^-]^2  = 9.1\times 10^{-6} = (X)(0.151 + 2X)^2\)

\(X= 2.08 \times 10^{-7}\)


\([Ca^{2+}]=X=2.08\times 10^{-7}\)

\(\%\ Ca^{2+}\ unprecipitated = \frac{2.08\times 10^{-7}\ M\ final}{0.0776\ M\ initial}\times 100 \%=  2.68\times 10^{-4}\ \% \ unprecipitated\)


Silver ion and bromine ion precipitate into \(AgBr_{(s)}\). The \(Ag^+\) concentration is about 0.065M If the \(Br^-\) concentration remains at \(1.0 \times 10^{17}\; M\), what will be the remaining of \([Ag^+]\) in solution when precipitation stops \((K_{sp}=7.7\times 10^{13})\)?


\[AgBr_{(s)} \rightleftharpoons Ag^+_{(aq)}+Br^-_{(aq)}\]


\(Q_{sp}=[Ag^+][Br^-]=(0.065)(1.0\times 10^{15})\)

\(Q_{sp}>K_{sp}\) therefore, precipitation occurs


\([Ag^+]\times (1.0\times 10^{15})=7.7\times 10^{13}\)

\([Ag^+]\ remaining=7.7\times 10^{-4}\ M\)


\([OH^-]\) is maintained at \(3.0 \times 10^{-2}\; M\) in a solution and \([Mg^{2+}] = 0.036\; M\). How much \(Mg^{2+}\) remain after \(Mg(OH)_2\) precipitate? (\(K_{sp} = 1.8 \times 10^{-11}\))


\([Mg^{2+}] = 2.0 \times 10^{-8}\; M\) remains.


If a constant \([F^-]=0.003\ M\) is maintained in a solution in which the initial \([Ca^{2+}] = 0.038\ M\), what percentage of \(Ca^{2+}\) will remain in solution after \(CaF_{2(s)}\) precipitates? What \([F^-]\) should be maintained to ensure only 3.4% of \(Ca^{2+}\) remains unprecipitated?


We first use the solubility product constant expression to determine \([Ca^{2+}]\) in a solution with 0.003 M \(F^-\)

\(K_{sp}= [Ca^{2+}][F^-]^2\)

\(5.3\times 10^{-9} = [Ca^{2+}](0.003)^2\)

\([Ca^{2+}]= \frac{5.3\times 10^{-9}}{(0.003)^2} = 5.89\times 10^{-4}\)

\(\% \ unprecipitated = \frac{5.89\times 10^{-4}\ M}{0.038\ M}\times 100\% = 1.5\%\)

Now, we want to find \([F^-]\) must be maintained to keep \([Ca^{2+}]_{final}\) =3.4%


\([Ca^{2+}]_{initial} = 0.034\times 0.038M = 1.29\times 10^{-3}\ M\)

\(K_{sp}= [Ca^{2+}][F^-]^2=5.3\times 10^{-9} = (1.29\times 10^{-3})[F^-]^2\)         

\([F^-]= \sqrt{\frac{5.3\times 10^{-9}}{1.29\times 10^{-3}}} = 0.0021\ M\) 


If \(K_{sp}\) of \(Zn(CN)_{2(s)}\) is \(3\times 10^{-16}\), what percentage of \(Zn^{2+}\) is precipitated when we know that

  • \([Zn^{2+}]_o= 0.36\;M\)
  • \([CN^-]_o= 5 \times 10^{-3}\;M\)
  • \(K_{sp}=3\times 10^{-16}\) for \(Zn(CN)_{2(s)}\).


\[Zn(CN)_{2(s)} \rightleftharpoons Zn^{2+}_{(aq)} + 2CN^-_{(aq)}\]

\([Zn^{2+}][CN]^2 =3\times 10^{-16}\)

\([Zn^{2+}]=\frac{3\times 10^{-16}}{(5\times 10^{-3})^2} = 1.2\times 10^{-11}\)

Due to the precipitation reaction, \([Zn]\) was reduced to \(3.33\times 10^{-9}\), signifying that it has essentially fully precipitated out of solution.


There are 320 g \(Na^+\) per kg of water

  1. should NaOH precipitate when \([OH^-] = 1.57\times 10^{-3}\ M\)?
  2. is it possible for \(Na\) and \(Cl\) to be separated in the water?


  1. no
  2. no


A sample of seawater contains 0.25 M \(Mg^{2+}\) and 0.0033 M \(Ca^{2+}\). Which will precipitate first when \(NaOH\) is added? What’s the concentration of \(X^{2+}\) ion remaining after \(X(OH)_2\) precipitate?


\(Mg(OH)_2\) will precipitate first, and it gets used almost completely.


0.01M of \(Cu^+\) and 0.01M of \(Ag^+\) are in solution. A piece of solid \(NaI\) is put in inside. What is the concentration of \(Ag^+\) as soon as \(CuI\) begins to precipitate?


\(Cu^+ + I^- \rightarrow CuI\) and \(K_{sp}\) is \(5.1\times 10^{-12}\)

\(Ag^+ + I^- \rightarrow AgI\) and \(K_{sp}\) is \(8.3\times 10^{-17}\)


\([Ag^+]= \frac{K_{sp}\ of\ AgI}{[I^-]\ in\ CuI}\)

\([Ag^+]= \frac{8.3 \times 10^{-17}}{5.1\times 10^{-10}}  = 1.627\times 10^{-7}\)


\(KI_{(aq)}\) is slowly added to a solution with \([Pb^{2+}] = [Ag^+] = 0.30\; M\). For \(PbI_2\), \(K_{sp} = 9.6 \times 10^{-9}\); for \(AgI\), \(K_{sp} = 6.2 \times 10^{-18}\).

  1. Which precipitate should form first, \(PbI_2\) or \(AgI\)?
  2. What is fractional precipitation?
  3. Can \(Pb^{2+}\)  and \(Ag^+\)  be effectively separated by fractional precipitation of their iodides?


  1. AgI will precipitate first because it has the smaller Ksp value.
  2. Fractional precipitation is the technique of separating two or more ions from a solution by adding a reactant that precipitates first one ion, then another, and so on.
  3. Yes because the ratio of the Ksp values compared to the molar concentration of .3 M for [Pb2+] and [Ag+] shows that both ions will completely dissociate.


If \(Cl^-\) is added to a solution that contains \(0.025\;M\) of \(Ag^+\) ions and \(0.00448\;M\) of \(Pb^{2+}\) ions. What concentration of \(Cl^-\) is required to precipitate each ion? \(K_{sp} (AgCl) = 1.8 \times 10^{-10}\) and \(K_{sp} (PbCl_2) = 1.6 \times 10^{-5}\).


  • After adding \(7.2 \times 10^{-9}\; M\) of \([Cl^-]\), silver will precipitate.
  • After adding \(3.57 \times 10^{-3}\; M\) of \([Cl^-]\), lead will precipitate.


0.23 M \(Na^+\) is added to a solution of 0.45 M \(Cl^-\) and 0.48 M \(OH^-\).

  1. Should \(NaCl\) or \(NaOH\) precipitate first?
  2. What concentration of \([OH^-]\) is required for the precipitation?
  3. Can these ions be effectively precipitated by fractional precipitation?


  1. NaCl
  2. 0.049 M
  3. yes


A concentration solution of potassium chromate is added to a solution that is 0.1 M \(Ba^{2+}\) and 0.1 M \(Sr^{2+}\)   Which ion will precipitate first? What is the percentage of this ion remaining when the other ion begins to precipitate?





\(BaCrO_{4(s)} \rightleftharpoons  Ba^{2+}+CrO_4^{2-}\)

\(K_{sp}=[Ba^{2+}][CrO_4^{2-}]\)\([CrO_4^{2-}]=1.2\times 10^{-9}\ M\)

\(1.2\times 10^{-10}=(0.1\ M)[CrO_4^{2-}]\)


\(SrCrO_{4(s)} \rightleftharpoons  Sr^{2+}+CrO_4^{-2}\)

\(K_{sp}=[Sr^{2+}][CrO_4^{2-}]\)\([CrO_4^{2-}]=3.5\times 10^{-4}\ M\)

\(3.5\times 10^{-5}=(0.1\ M)[CrO_4^{2-}]\)



\(BaCrO_{4(s)} \rightleftharpoons  Ba^{2+}+CrO_4^{2-}\)


\(1.2\times 10^{-10}=[Ba^{2+}][3.5\times 10^{-4}]\)

\([Ba^{2+}]=3.4\times 10^{-7}\ M\)

\(\% \ Ba^{2+}\ remaining= \frac{3.4\times 10^{-7}\ M}{0.1}\times 100\% = 0.000343\% \)


0.4 M \(Cu^{2+}\) is added to a solution of 0.039 M \(CO_2^3\) and 0.039 M \(Cl^-\).

  1. Which precipitates first?
  2. When the second ion starts to precipitate, what is the concentration of the first ion?
  3. Can these be effectively separated by fractional precipitation?


  1. \(CuCO_3\)
  2. 0.028 M
  3. yes


An aqueous solution that 2.00 M in \(Cu(CO_3)\) is slowly added from a buret to a aqueous solution that is 0.020 M in \(Cl^-\)  and 0.30 M in \(Br^-\):

  1. Which ion will precipitate first?
  2. What is the remaining concentration of the first ion when the second ion begins to precipitate?
  3. Is the separation of \(F^-\) and \(Br^-\) feasible by fractional precipitation in this solution?


From the table of \(K_{sp}\) values (Table E3):

\[K_{sp} \; (CuCl)= 1.7 \times 10^{-7}\]

\[K_{sp}\, (CuBr) = 6.3 \times 10^{-9}\]

\(CuBr\) will be the first to precipitate due to its smaller \(K_{sp}\) value.

\(CuCl\) will precipitate when \(Q_{sp} > K_{sp}\). First find the \(Cu\) concentration when \(CuCl\) first precipitates. 

\[1.7 \times 10^{-7}= 0.02 [Cu^+]\]

\[[Cu^+]=8.5 \times 10^{-7}\]

\[6.3 \times 10^{-9}=8.5 \times 10^{-7} [Br^-]\]

\[0.007412 = [Br^-]\]

\([Br^-]\) drop from 0.3 M to 0.007412 M before \(CuCl\) precipitate begins to form. This means that  \frac{0.007412}{0.3}= 0.0247= 2.47\% \). This small percentage of \(Br^-\) ions that remain unreacted in the solution implies that fractional precipitation is feasible.


If a solution contains \(5.2 \times 10^{-3} \; M\) of \(Pb^{2+}\) and \(4.8 \times 10^{-2} \; M\) of \(Cu^+\). What concentration of \(I^-\) is needed to be added to start a precipitation reaction?


\(1.10\times 10^{-10}\ M\) of \(I^-\) is needed to begin precipitation.


a) Can \(NaOH\) be separated by precipitating \(NaCl\) but not \(HCl\) if all concentrations are 0.2 M?

b) Will the separation be complete?


a) Yes

b) No


Which of the following solids are more soluble in an acidic solution than in pure water: \(NaCl\), \(Fe(OH)_2\), \(CaF_2\), \(ZnCO_3\), or \(C_5H_4CO_2H\)?


  • \(NaCl\): They will have similar solubilities in both solvents because \(NaCl\) does not produce any ions that will directly react with an acid.
  • \(Fe(OH)_2\): This will be more soluble in an acidic solution because the \(OH^-\) ions created will form water with \(H^+\) ions present in the acidic solution.
  • \(CaF_2\): This will be more soluble in an acidic solution because the \(F^-\) ions created will react with the \(H^+\) ions present in the acidic solution to increase the pH.
  • \(ZnCO_3\): This will be more soluble in an acidic solution because the \(CO_3\) will lose an oxygen to form water with 2 protons already present in the acidic solution.
  • \(C_5H_4CO_2H\): This will be more soluble in pure water because in this molecule there are \(H_3O^+\) ions that will not react with the acidic solution at all, and will therefore be more soluble in water.


Is \(Ce(IO_3)­_3\) able to precipitate from a solution that contains 500 mL of 0.10 M \(KIO_3\) and 200 mL of 0.05 M \(Ce(NO_3)_3\)? (\(K_{sp} = 1.9 \times 10^{-10}\))


It’s going to precipitate.


Can \(Pb(OH)_2\) precipitate from a solution that contains \(3.2 \times 10^{-3}\; M \;PbCl_2\) and \(0.05\; M \;NH_3\)?


No precipitation would occur.


A saturated solution of \(Ca_3(PO_4)_2\) has \([Ca^{2+}] = [PO_4^{3-}] = 2.9\times 10^{-7}\ M\). Will \(Ca_3(PO_4)_2\) precipitate?




The solubility of \(Ba(OH)_2\) in a particular buffer solution is \(0.5\frac{g}{L}\). What is the pH of the buffer solution? \(K_{sp}= 5.0 \times 10^{-3}\)


First find \([Ba^{2+}]\) using \(0.5\frac{g}{L}\)

\[\dfrac{0.5\; grams}{171.327\; \frac{g}{mol}} = 0.00584\; \frac{mol}{L}= [Ba^{2+}]\]

\[5 \times 10^{-3}=[Ba^{2+}][OH^-]^2\]


\[pOH= -log(0.925)= 0.034\]

\[pH= 13.9663\]


The solubility of \(Na^{2+}\) in 2 M \(CaCl\) is \(1.3\times 10^{-2} \frac{mol}{L}\). If \(K_{sp}\) of \(Na^{2+}\) is \(3.1\times 10^{-10}\), what is the value of \(K_f\) for \(NaCl\)?


\(K_f=2.88\times 10^{-2}\)


If a 0.20 mol sample of \(Zn(NH_3)_4SO_4\) is dissolved in 1.00L of an aqueous solution of 1.0 M \(NH_3\), what’s the concentration of \([Zn^{2+}]\)? Hint: Use equilibrium constants in Table E4.


Many complex ions such as tetrammine zinc (II), \(Zn(NH_3)_4^{+2}\), are loosely aggregated and tend to dissociate in a water solution until an equilibrium is established between the complex ion and its components.

\[Zn(NH_3)_4^{+2} \rightleftharpoons Zn^{2+} + 4 NH_3\]

\[K= \dfrac{[Zn^{2+}][NH_3]^4}{[Zn(NH_3)^{+2}]^4} = 7.8 \times 10^8\]

Video Solution


The solubility of \(AlF_{3(s)}\) in 1.5 M \(OH^{-}_{(aq)}\) is \(2.0\times 10^{-4}\frac{mol}{L}\). Given that \(K_{sp}\) of \(AlF_3\) is \(3.7 \times 10^{-9}\), what is the value of \(K_f\) for \(Al(OH)_3\)?


To get the net reaction we combine the solubility product expression for \(AlH_{3(s)}\) with the formation expression of \(Al(OH)_{3(s)}\).


Solubility: \(AlF_{3(s)} \rightleftharpoons Al^{3+}_{(aq)}+3F^-_{(aq)}\) , \(K_{sp}=3.7\times 10^{-9}\)

Formation: \(Al^{3+}_{(aq)} + 3OH^-_{(aq)} \rightleftharpoons Al(OH)_{3(aq)}\) , \(K_f = ?\)

Net Reaction: \(AlF_{3(s)}+3OH^-_{(aq)} \rightleftharpoons Al(OH)_{3(aq)}+3F^-_{(aq)}\) ,  \(K_{overall}=K_{sp}K_f\)


\(K_{overall}=\frac{[Al(OH)_3 ][F^-]^3}{[OH^-]^3} =\frac{(2.0\times 10^{-4})^4}{(1.5)^3} =4.74\times 10^{-16}=K_f\times  (3.7\times 10^{-9})\)

\(K_f=  \frac{ 4.74\times 10^{-16} }{ 3.7\times 10^{-9} } =1.27\times 10^{-7}\)


A mixture of \(PbSO_4(s)\) and \(PbS_2O_3(s)\) is shaken with pure water until a saturated solution is formed. Both solids remain in excess. What is \([Pb^{2+}]\) in the saturated solution? For \(PbSO_4\), \(K_{sp} = 3.4 \times 10^{-7}\); for \(PbS_2O_3\), \(K_{sp} = 7.6 \times 10^{-4}\).


\(K_{sp}(PbSO_4) = [Pb^{2+}][SO_4^{2-}]\) with \([Pb^{2+}] = [SO_4^{2-}]\)

      So \(K_{sp}(PbSO_4)= [Pb^{2+}]^2\)

\(K_{sp}(PbS_2O_3) = [Pb^{2+}][S_2O_3^{2-}]\) with \([Pb^{2+}] = [S_2O_3^{2-}]\)

     So \(K_{sp}(PbS_2O_3) = [Pb^{2+}]^2\)

\(K_{sp}(PbSO4) \times K_{sp}(PbS_2O_3) = [Pb^{2+}]^4\)

\((3.4\times 10^{-7})(7.6\times 10^{-4}) = [Pb^{2+}]^4\)

\(\mathbf{[Pb^{2+}] = .0040 M}\)


A mixture of \(AlPO_4(s)\) and \(AlPO_3(s)\) is shaken with water until a saturated solution is formed. Both of the solids remain in excess. What is \([Al^{3+}]\) in the saturated solution? For \(AlPO_4\), \(K_{sp} = 1.3 \times 10^{-6}\) and for \(AlPO_3\), \(K_{sp} = 3.5 \times 10^{-4}\).


There are three relationships you need for this question:

\(1.3 \times 10^{-6}=[Al^{3+} ][PO_4^{3-}]\)

\(3.5 \times 10^{-4}=[Al^{3+} ][PO_3^{3-} ]\)

The third is the electro neutrality equation, which states that the total positive charge concentration must equal the total negative charge concentration.

\([Al^{3+} ]=[PO_4^{3-} ]+[PO_3^{3-}]\)

Use the first two expressions to solve for the concentration of each anion, then substitute them into the electroneutrality expression and solve for \([Al^{3+}]\).

\([PO_4^{3-}]=\frac{1.3 \times 10^{-6}}{[Al^{3+}]}\)       

\([PO_3^{3-}]=\frac{3.5 \times 10^{-4}}{[Al^{3+}]}\)      

\([Al^{3+}]=\frac{1.3\times 10^{-6}}{[Al^{3+}]}+\frac{3.5 \times 10^{-4}}{[Al^{3+}]}\)

\([Al^{3+} ]^2= 1.3\times 10^{-6}+3.5\times 10^{-4}=3.5\times 10^{-4}\)          

\([Al^{3+}]= \sqrt{(3.5\times 10^{-4}}=0.019\ M \)


A mixture of \(CaSO_4\) and \(AgSO_4\) is shaken with water until the solution is saturated. What is \([Ag^{2+}]\)?


0.0314 M

Q17.83 - (Answer is not given)

At what pH does \(Cr(OH)_3\) just starts to precipitate from an 0.10 M \(Cr^{3+}\) solution? From this calculation justify why it is impossible to prepare an 0.10 M \(Cr^{3+}\) solution at pH=7. The \(K_{sp}\) of  \(Cr(OH)_3\) is \(6.3 \times 10^{-31}\).