Question:
Saccharin (\(K_a = 2 \times 10^{-12} \)) is a weak acid represented by formula HSaC. A \(4 \times 10^{-4}\) mole amount of saccharin is dissolved in 200mL of water of pH 3. Assuming no change in volume, calculate the concentration of \(SaC^{-1}\) ions in the resulting solution at equilibrium.
Line of Thinking:
At equilibrium, saccharin will dissociate to give \(SaC^{-1}\) ions and \(H^+\) ions. pH of the solution is 3 and hence we will be able to calculate the concentration of \(H^+\) ions in solution. The initial concentration of saccharin can also be calculated from the data given in the question.
Also, \(K_a = \dfrac {[Sac^{-1}]\times[H^+]}{[HSaC]} = \dfrac{C\alpha \times [H^+]}{C(1 - \alpha)}\) where \(\alpha\) is the degree of dissociation and \(C\) is the initial concentration of saccharin.
Can you calculate the answer based on the above Line of Thinking (LoT)?
Can you solve this question based on the lines of thinking given?
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