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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?

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.

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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 !!

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.

We know that:

(a) $\ce{HB_{(aq)} <-> H^{+}_{(aq)} + B−_{(aq)}}$

(b) $\frac{[H+][B−]}{[HB]}=K_a=6.46∗10−5$

(c) pH for pure water is 7

So rearrange (b) and plug in our assumptions

$ [HB] = \dfrac{[H+][B−]}{6.46∗10^{−5}} = \dfrac{(1*10^{-7})(5.0*10^{-7})}{6.46∗10^{−5}} = 7.7 * 10^{-10}$

Since $7.7 * 10^{-10} << 5.0 * 10^{-7} $ we can safely ignore protonation of $\ce{B^-}$ to $\ce{HB}$.

(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 !!

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

$[\ce{Ag^+}] [\ce{B^{-}}] = (1.66 * 10^{-6})(1.51 * 10^{-7}) = 2.51 * 10^{-13}$

(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 $

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  • $\begingroup$ Great ! But why would I want to check if [B-] >> [HB] ? And , in both the cases , why would I want to add the concentrations of [B-] and [HB] ? Aren't they two different things ? $\endgroup$ – user22729 Nov 10 '15 at 6:34
  • $\begingroup$ In the first solution in pure water you want to check for the protonation of B− to HB. That would cause more silver benzoate to dissolve as it does in the buffer solution. There is "some" HB in pure water but the quantity is minuscule to the B- species. $\endgroup$ – MaxW Nov 10 '15 at 6:43
  • $\begingroup$ Re adding [B-] and [HB] - There were no B species in solution before the solid sodium benzoate was added. So $\ce{[Ag^{+}] = [B^{-}] + [HB]}$. in other words all the B species had to come from silver benzoate that dissolved. $\endgroup$ – MaxW Nov 10 '15 at 6:46
  • $\begingroup$ So we are adding them because both of them are B species. And in the first case as [HB] is negligible , the concentration of [B-] + [HB] is considered to be the concentration of [B-] alone , but in the second case , as [HB] is not negligible , we have to consider it in the total concentration [B-] + [HB]...Right ? $\endgroup$ – user22729 Nov 10 '15 at 6:50
  • $\begingroup$ Yes. Both [B−] and [HB] are "b species" which were from silver benzoate. $\endgroup$ – MaxW Nov 10 '15 at 6:52
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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

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  • $\begingroup$ Thanks ! I have 2 questions , the answers of which will for sure solve my problem. 1)What is [Ag+]i and [BzO-]i ? 2)For the expression of [BzO-]i , why have you added [BzOH] and [BzO-] ? $\endgroup$ – user22729 Nov 10 '15 at 6:23
  • $\begingroup$ They are the initial quantity. And for the second question it is because we need to take all the species containing BzO- :) $\endgroup$ – ParaH2 Nov 10 '15 at 9:07
  • $\begingroup$ Just for clarification , BzO- is benozate ion , right ? $\endgroup$ – user22729 Nov 10 '15 at 13:27
  • $\begingroup$ Yes of course ;) $\endgroup$ – ParaH2 Nov 10 '15 at 13:43
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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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  • $\begingroup$ Why have you added the concentrations of Benzoate Ion and Benzoic acid in your expression for 's' ? $\endgroup$ – user22729 Nov 10 '15 at 6:15
  • $\begingroup$ Of course, when bonzoate ion is in aqueous solution, it reacts with water to form its conjugated acid. This reaction is not quantitative. It's an equilibrium. So, the initial concentration of bonzoate ion, i.e. (s) becomes the sum of the remaining concentration of bonzoate ion and the concentration of the formed benzoic acid. This equation is also known under the name: conservation of matter (benzoic ion here). $\endgroup$ – Yomen Atassi Nov 10 '15 at 7:55

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