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Question: In a 0.1 M solution, a monobasic acid is \[1\% \] ionized. The ionization constant of the acid is: ...

In a 0.1 M solution, a monobasic acid is 1%1\% ionized. The ionization constant of the acid is:
A. 1×10−31 \times {10^{ - 3}}
B. 1×10−71 \times {10^{ - 7}}
C. 1×10−51 \times {10^{ - 5}}
D. 1×10−141 \times {10^{ - 14}}

Explanation

Solution

When acids are placed in an aqueous medium or are interacted with basic substances, they tend to release a hydrogen atom from the molecule of the acid. The degree and ease to which this hydrogen atom gets released from the molecule of the acid determine the acidity of the acid. The higher the dissociation of the hydrogen atom, the higher is the acidity of that particular acid.

Complete step by step answer:
In the case of monobasic weak acids, ionization tendency is very low, as a result, the molar concentration of hydrogen ion cannot be equal to the concentration of the acid. In this case, a new term is introduced which is ionization constant (Ka)\left( {{K_a}} \right) . The ionization constant of weak acid tells us that weak acids dissociate partly in a dilute solution. Based on the concentration of the acid, we can determine the ionization constant. The ionization of monobasic acid is shown below,

C(1 - \alpha )\hspace{3em} \qquad C\alpha \hspace{3em} C\alpha \\\ $$ Now let the degree of ionization of weak monobasic acid is $$\alpha $$ . Therefore, the concentration of the hydrogen ion of weak monobasic acid is Ca. Where C is 0.1M. Now when a monobasic acid is $$1\% $$ ionized. Therefore, the degree of ionization is, $$\alpha = \dfrac{1}{{100}}$$ So, the concentration of hydrogen ion is $$C\alpha = \dfrac{1}{{100}} \times 0.1$$ The ionization constant is the ratio of the concentration of ionized ions with unionized acids. Shown below, $$ {K_a} = \dfrac{{C\alpha \times C\alpha }}{{C(1 - \alpha )}} \\\ $$ $\Rightarrow$ $$ {K_a} = \dfrac{{C{\alpha ^2}}}{{(1 - \alpha )}} \\\ $$ Now put the values as follows and find out the value of ionization constant, $$ {K_a} = \dfrac{{C{\alpha ^2}}}{{(1 - \alpha )}} \\\ $$ $\Rightarrow$ $$ {K_a} = \dfrac{{0.1{{\left( {\dfrac{1}{{100}}} \right)}^2}}}{{(1 - \dfrac{1}{{100}})}} \\\ $$ $\Rightarrow$ $$ {K_a} = \dfrac{{{{10}^{ - 5}}}}{{.99}} \\\ $$ $\Rightarrow$ $$ {K_a} = \dfrac{{{{10}^{ - 5}}}}{1} \\\ $$ $\therefore$ $$ {K_a} = {10^{ - 5}} \\\ $$ **So, the correct answer is C.** **Note:** This property of acidity is mathematically measured using a method known as the pH scale. Depending on the degree of dissociation of hydrogen atoms, the acidity of the substance is calculated. On a scale of 1 to 14, substances with a pH value between 1 to 7 are known as acids, with 1 being the most acidic and 7 being the least acidic.