Question

What would be \[[{H^ + }]\] of 0.006 M benzoic acid (\[{K_a} = 6 \times {10^{ - 5}}\] )
A.\[0.6 \times {10^{ - 4}}\] M
B.\[6 \times {10^{ - 4}}\] M
C.\[6 \times {10^{ - 3}}\] M
D.\[3.6 \times {10^{ - 4}}\] M

Answer 1

Hint: We know that \[{K_a}\] is the dissociation constant of a compound can be expressed by the given expression
\[ {K_a} = \dfrac{{[{H^ + }][{C_6}{H_5}CO{O^ - }]}}{{[{C_6}{H_5}COOH]}}\]
Where [ ] indicates the concentration of the given term in moles per litre
It takes into account the concentration of \[[{H^ + }]\] ions and thus we can get a relationship for the dissociation constant and the concentration of \[[{H^ + }]\] ions and the molarity of the acid.

Complete answer:
Let us take a look at the given equation
\[ \Rightarrow {K_a} = \dfrac{{[{H^ + }][{C_6}{H_5}CO{O^ - }]}}{{[{C_6}{H_5}COOH]}}\]
For weak acids such as benzoic acid we can say that
\[ \Rightarrow [{H^ + }] = [{C_6}{H_5}CO{O^ - }] = x\]
Therefore we can rewrite the equation and find that
\[ \Rightarrow {K_a} = \dfrac{{{x^2}}}{{[{C_6}{H_5}COOH]}}\]
Where x is the concentration of \[[{H^ + }]\] ions
\[ \Rightarrow [{H^ + }] = \sqrt {[{C_6}{H_5}COOH] \times {K_a}} \]
From this equation we can say that molarity of the acid solution is given in the question as 0.006 M
The dissociation constant of benzoic acid is given as \[{K_a} = 6 \times {10^{ - 5}}\]
This by substituting in the given equation we get
\[ \Rightarrow [{H^ + }] = \sqrt {0.006 \times 6 \times {{10}^{ - 5}}} \]
\[ \Rightarrow [{H^ + }] = 6 \times {10^{ - 4}}M\]
Thus we can say that the concentration of \[[{H^ + }]\] ions in the given solution of benzoic acid is \[6 \times {10^{ - 4}}M\]
Thus we can say that option (B) is the correct answer for the given question.

Note:
We all know that \[pH\] is defined as the \[ - \log [{H^ + }]\] of a given solution and this value which is between 1-14 will tell us about the acidity of the given solution.
So by knowing the concentration of the \[[{H^ + }]\] ion in the question we can calculate the pH of the solution.
\[pH\] of this solution is given by
\[ \Rightarrow pH = - \log [6 \times {10^{ - 4}}]\]
So by finding out the value we can say that
\[pH = 3.22\]
This is the \[pH\] of the given solution and we can say it is acidic.