Solubility Product and Buffers


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> The ionization constant of benzoic acid is $6.46 \times 10^{-5}$ and solubility product for silver benzoate is $2.5 \times 10^{-13}$. How many times silver benzoate is more soluble in a buffer of $\mathrm{pH}$ 3.19 compared to its solubility in pure water?
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Equation used to calculate solubility of silver benzoate in water:


$$\ce{C6H5COOAg <=> C6H5COO- + Ag+}$$


I've calculated the solubility of silver benzoate in water , it is $5 \times 10^{-7} \pu{mol/L}$ (since that is all I calculated, this is the only equilibrium equation I know).


But I cannot understand what to do next.




Ok, the answer is in three parts. Let's use $B^-$ to represent the benzoate anion. 


**(1) Solubility of Silver Benzoate in pure Water**


$\ce{AgB\_{(s)} <-> Ag+\_{(aq)} + B^−\_{(aq)}} \quad\quad K\_{sp} = 2.5 \* 10^{-13}$


thus:
$\ce{[Ag^+] [B^-]} = 2.5 \* 10^{-13}$


assume $\ce{[Ag^+] = [B^{-}]}$ then


$[Ag^+] = [B^-] = \sqrt{2.5 \* 10^{-13}} \text{m/l} = 5.0 \* 10^{-7} \text{m/l}$


**!! Check !!**



> 
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> > We want to check if $\ce{[B^{-}] >> [HB]}$ so that $[B^{−}\_{(aq)}] + [HB\_{(aq)}] \approx 5.0 \* 10^{-7}$ and that the pH will stay at 7.
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> > We know that:
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> > (a) $\ce{HB\_{(aq)} <-> H^{+}\_{(aq)} + B−\_{(aq)}}$
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> > (b) $\frac{[H+][B−]}{[HB]}=K\_a=6.46∗10−5$
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> > (c) pH for pure water is 7
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> > So rearrange (b) and plug in our assumptions
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> > $ [HB] = \dfrac{[H+][B−]}{6.46∗10^{−5}} = \dfrac{(1\*10^{-7})(5.0\*10^{-7})}{6.46∗10^{−5}} = 7.7 \* 10^{-10}$
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> > Since $7.7 \* 10^{-10} << 5.0 \* 10^{-7} $ we can safely ignore protonation of $\ce{B^-}$ to $\ce{HB}$. 
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**(2)Solubility of Silver Benzoate in a buffer of pH 3.19** 


(a) $\frac{[H+][B−]}{[HB]}=K\_a=6.46∗10^{−5}$


(b) pH for buffer is 3.19 therefore $\ce{[H^+]}$ = 6.45 \* $10^{-4}$


(c) $\ce{AgB\_{(s)} <-> Ag+\_{(aq)} + B^−\_{(aq)}} \quad\quad K\_{sp} = 2.5 \* 10^{-13}$


So rearrange (a) and plug in $\ce{[H^+]}$


$\dfrac{[B−]}{[HB]}= \dfrac{6.46∗10^{−5}}{6.45 \* 10^{-4}} = 0.10 $


Thus in this acid solution we must consider the protonation of $\ce{B^{-}}$ to $\ce{HB}$ when solving for the solubility of silver benzoate. 


so we another equation.


(d) $\ce{AgB\_{(s)} <-> Ag+\_{(aq)} + B^{−}\_{(aq)} + HB\_{(aq)}}$


but we know that $\ce{[Ag+\_{(aq)}] = [B^{−}\_{(aq)}] + [HB\_{(aq)}] = [B^{−}\_{(aq)}] (1 + $\dfrac{\ce{[HB\_{(aq)}]}}{\ce{[B^{−}\_{(aq)}]}}$) = 11 [B^{−}\_{(aq)}]}$


so from (c) we get:


$\ce{11[B^{−}]^2} = 2.5 \* 10^{-13}$


$\ce{[B^{−}]}^2 = \dfrac{2.5 \* 10^{-13}}{11} = 2.27 \* 10^{-14}$


$\ce{[B^{−}]} = 1.51 \* 10^{-7}$


and thus


$\ce{[HB]} = 10 \ce{[B^{−}]} = 1.51 \* 10^{-6}$


$\ce{[Ag^+]} = 11 \ce{[B^{-}]} = 11 \* 1.51 \* 10^{-7} = 1.66 \* 10^{-6}$


**!! CHECK !!**



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> > 
> > $[\ce{HB}] + \ce{[B^{-}]} = 1.51 \* 10^{-6} + 1.51 \* 10^{-7} = 1.66 \* 10^{-6}$
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> > $[\ce{Ag^+}] [\ce{B^{-}}] = (1.66 \* 10^{-6})(1.51 \* 10^{-7}) = 2.51 \* 10^{-13}$ 
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**(3) Ratio of Solubilities of Silver Benzoate**


$\text{Ratio} = \dfrac{\text{solubility in buffer}}{\text{solubility in water}} = \dfrac{1.66 \* 10^{-6}}{5.0 \* 10^{-7}} = 3.3 $


 6

In fact its "as easy as pie". But I know its always easy to think about things like it was an evidence. I will give my approach to solve this kind of problems. Obviously you can do the same with complexation, liquid/liquid extraction and so on. 


Fisrt you have $$\ce{BzOAg}\leftrightharpoons \ce{BzO^-}+\ce{Ag^+} \text{ with the constant }\ce{K\_s}$$


But you also have a second possible reaction which is $$\ce{BzO^-}+\ce{H\_2O}\leftrightharpoons \ce{BzOH}+\ce{HO^-} \text{ with the constante }\ce{K\_b}$$


You know that $$\ce{K\_s}=[\ce{Ag^+}]\_i[\ce{BzO^-}]\_i$$


At any time $[\ce{Ag^+}]\_i=[\ce{Ag^+}]$ and 


$$\begin{align}[\ce{BzO^-}]\_i&=[\ce{BzOH}]+[\ce{BzO^-}]\\&=[\ce{BzO^-}]\left(1+\frac{[\ce{H\_3O^+}]}{\ce{K\_a}}\right)\end{align}$$


Then you have $$\ce{s^2}=\ce{K\_s}\times \left(1+\frac{[\ce{H\_3O^+}]}{\ce{K\_a}}\right)$$


Now you also know that $[\ce{H\_3O^+}]=\ce{10^{-pH}}$ you have:


$$\ce{s}=\sqrt{\ce{K\_s}\times\left(1+\frac{10^{-pH}}{\ce{K\_a}}\right)}$$


I hope it will help you. Ask me if you have questions. Have a nice day.


Friendly


 1

As ion benzoate is a base in the couple $\ce{C6H5COOH /C6H5COO- }$, we can write
$$s=\ce{[Ag+] =[C6H5COO- ] + [C6H5COOH ]}$$
$$s=\ce{[Ag+] =[C6H5COO- ] (1+\frac{[C6H5COOH ]}{[C6H5COO- ] })}$$


$$s=\ce{[Ag+] =[C6H5COO- ]} (1+\frac{[\ce{H3O+} ]}{K\_\mathrm{a} })$$
On the other hand,$$K\_\mathrm{sp}=\ce{[Ag+] \times [C6H5COO- ]} $$
$$K\_\mathrm{sp}=\frac{s^2}{(1+\frac{[\ce{H3O+} ]}{K\_\mathrm{a} })} $$
$$s=\sqrt{K\_\mathrm{sp}} \times \sqrt{{(1+\frac{[\ce{H3O+} ]}{K\_\mathrm{a} })}} $$


So, the solubility in the acidic buffer solution is increased by the factor:$\sqrt{1+\frac{[\ce{H3O+} ]}{K\_\mathrm{a} }}$


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