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Question: What would be \[[{H^ + }]\] of 0.006 M benzoic acid (\[{K_a} = 6 \times {10^{ - 5}}\] ) A.\[0.6 \t...

What would be [H+][{H^ + }] of 0.006 M benzoic acid (Ka=6×10−5{K_a} = 6 \times {10^{ - 5}} )
A.0.6×10−40.6 \times {10^{ - 4}} M
B.6×10−46 \times {10^{ - 4}} M
C.6×10−36 \times {10^{ - 3}} M
D.3.6×10−43.6 \times {10^{ - 4}} M

Explanation

Solution

We know that Ka{K_a} is the dissociation constant of a compound can be expressed by the given expression
Ka=[H+][C6H5COO−][C6H5COOH]{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+][{H^ + }] ions and thus we can get a relationship for the dissociation constant and the concentration of [H+][{H^ + }] ions and the molarity of the acid.

Complete answer:
Let us take a look at the given equation
⇒Ka=[H+][C6H5COO−][C6H5COOH]\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
⇒[H+]=[C6H5COO−]=x\Rightarrow [{H^ + }] = [{C_6}{H_5}CO{O^ - }] = x
Therefore we can rewrite the equation and find that
⇒Ka=x2[C6H5COOH]\Rightarrow {K_a} = \dfrac{{{x^2}}}{{[{C_6}{H_5}COOH]}}
Where x is the concentration of [H+][{H^ + }] ions
⇒[H+]=[C6H5COOH]×Ka\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 Ka=6×10−5{K_a} = 6 \times {10^{ - 5}}
This by substituting in the given equation we get
⇒[H+]=0.006×6×10−5\Rightarrow [{H^ + }] = \sqrt {0.006 \times 6 \times {{10}^{ - 5}}}
⇒[H+]=6×10−4M\Rightarrow [{H^ + }] = 6 \times {10^{ - 4}}M
Thus we can say that the concentration of [H+][{H^ + }] ions in the given solution of benzoic acid is 6×10−4M6 \times {10^{ - 4}}M
Thus we can say that option (B) is the correct answer for the given question.

Note:
We all know that pHpH is defined as the −log⁡[H+] - \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+][{H^ + }] ion in the question we can calculate the pH of the solution.
pHpH of this solution is given by
⇒pH=−log⁡[6×10−4]\Rightarrow pH = - \log [6 \times {10^{ - 4}}]
So by finding out the value we can say that
pH=3.22pH = 3.22
This is the pHpH of the given solution and we can say it is acidic.