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Question: Saccharin \(({{K}_{a}}=2\times {{10}^{-12}})\) is a weak acid represented by formula HSac. A \(4\tim...

Saccharin (Ka=2×10−12)({{K}_{a}}=2\times {{10}^{-12}}) is a weak acid represented by formula HSac. A 4×10−44\times {{10}^{-4}} mole amount of saccharin is dissolved in 200 cm3c{{m}^{3}} water of pH=3. Assuming no change in the volume, calculate the concentration ofSac−Sa{{c}^{-}} ions in the resulting solution at equilibrium.

Explanation

Solution

The answer to this question includes the calculation of concentration of saccharin and then the dissociation taking place in the presence of H+{{H}^{+}} ions that is calculated using pH and also includes the concept of the common ion effect.

Complete step – by – step answer:
We have come across the chapters in chemistry that deal with the calculation of various parameters such as concentration, molarity, molality, normality, mole fraction etc.
Now, we shall reconsider these calculations which also include the pH calculation.
Now saccharin concentration according to the data, can be calculated as,
[HSac]=molesvolume(l)=4×10−4200/1000  =2×10−3M[HSac]=\dfrac{moles}{volume(l)}=\dfrac{4\times {{10}^{-4}}}{{200}/{1000}\;}=2\times {{10}^{-3}}M
Since, pH = 3 and by the formula of pH we have,
pH = - log[H+][{{H}^{+}}]
⇒\Rightarrow 3 = - log[H+][{{H}^{+}}]
⇒\Rightarrow [H+][{{H}^{+}}] = 10−3{{10}^{-3}}
Now, the dissociation of HSac takes place in presence of [H+]=10−3[{{H}^{+}}]={{10}^{-3}}
Now, dissociation of HSac is as shown below,
HSacH++Sac−HSac{{H}^{+}}+Sa{{c}^{-}}
Now, before dissociation, the concentration of HSac is2×10−32\times {{10}^{-3}}and concentration of [H+][{{H}^{+}}]is 10−3{{10}^{-3}}and that of Sac−Sa{{c}^{-}}is 0.
This shows that in presence of H+{{H}^{+}}, the dissociation of saccharin is negligible and this is due to the common ion effect and because of this at equilibrium time the concentration of HSac and [H+][{{H}^{+}}] can be written as,
Ka=[H+][Sac−][HSac]{{K}_{a}}=\dfrac{[{{H}^{+}}][Sa{{c}^{-}}]}{[HSac]}
⇒2×10−12=10−3×[Sac−]2×10−3\Rightarrow 2\times {{10}^{-12}}=\dfrac{{{10}^{-3}}\times [Sa{{c}^{-}}]}{2\times {{10}^{-3}}}
Therefore, [Sac−]=4×10−12[Sa{{c}^{-}}]=4\times {{10}^{-12}}

Therefore, at constant volume the concentration of Sac−Sa{{c}^{-}} ions will be 4×10−124\times {{10}^{-12}}.

Note: Note that Saccharin is also called by the name sodium saccharin which is used as an artificial sweetener with no food energy and it has about 300 – 400 times as sweet as that of sucrose but gets bitter or metallic aftertaste which is especially at higher concentrations. It is used to sweeten the beverages, candies, cookies etc.