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Question: Calculate the percentage ionization of 0.01 M acetic acid in 0.1 M \(1.8\times {{10}^{-5}}\) \(HCl\)...

Calculate the percentage ionization of 0.01 M acetic acid in 0.1 M 1.8×10−51.8\times {{10}^{-5}} HClHCl. Ka{{K}_{a}}of acetic acid is 1.8×10−51.8\times {{10}^{-5}}.
A. 0.18%
B. 0.018%
C. 1.8%
D. 18%

Explanation

Solution

pH is a measure of hydrogen ion concentration in an aqueous solution. The pH scale ranges from 0 to 14 where the values less than 7 indicates acidity while pH exactly equal to 7 is neutral and a high value of pH value indicates alkalinity.

Complete step by step solution: Other than pH value there are some other terms Ka,pKa,Kb,pKb{{K}_{a}},p{{K}_{a}},{{K}_{b}},p{{K}_{b}} which are also helpful to predict whether a species will donate or accept protons at a specific pH value. These terms describe the degree of ionization of an acid or base and are said to be true indicators of acid or base strength because adding water to a solution will not change the equilibrium constant values Ka{{K}_{a}} Ka,pKa{{K}_{a}},p{{K}_{a}}relate to acids while Kb,pKb{{K}_{b}},p{{K}_{b}}deal with bases.
[H+]=10−1[{{H}^{+}}]={{10}^{-1}}

| CH3COOH⇌C{{H}_{3}}COOH\rightleftharpoons | CH3COO−+H+C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}
---|---|---
Initially| 0.01 M| 0                                  00\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0
After dissociation| 0.01-x | x                                  x+10−1x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x + {{10}^{-1}}

Ka=[CH3COOH−][H+][CH3COOH]{{K}_{a}}=\dfrac{[C{{H}_{3}}COO{{H}^{-}}][{{H}^{+}}]}{[C{{H}_{3}}COOH]}
[x][x+10−1][0.01−x]=1.8×10−5\dfrac{[x][x+{{10}^{-1}}]}{[0.01-x]}=1.8\times {{10}^{-5}}
Now we can say that the acid is weak so the value of 0.01−x0.01-xwould be equal to 0.01 and also x+0.1x+0.1is 0.1
∴Ka=x×0.10.01=1.8×10−5\therefore {{K}_{a}}=\dfrac{x\times 0.1}{0.01}=1.8\times {{10}^{-5}}
∴x=1.8×10−6\therefore x=1.8\times {{10}^{-6}}
Therefore % ionization = x×1000.01=1.8×10−60.01×100=0.018\dfrac{x\times 100}{0.01}=\dfrac{1.8\times {{10}^{-6}}}{0.01}\times 100=0.018%
Hence we can say that option B is the correct answer.
Note: A large value of Ka{{K}_{a}} indicates that the given compound is strong acid which means that the acid is largely dissociated into its ions and it also means the formation of products in the reaction is favored whereas small value means weak acid.