Question

An acid solution of pH 6 is diluted hundred times. The pH of resulting solution approximately becomes
(A) 6.95
(B) 6
(C) 4
(D) 9

Answer 1

Hint pH is the negative logarithm of hydrogen ion. The pH of a solution is the numerical value of the negative power of ten that rose to express the hydrogen ion concentration. An aqueous solution having pH value less than 7 will be acidic in nature, and if more than 7 it will be basic in nature. The pH value of water is 7. Nature of water is natural, for water $[Hh+]=[O{{H}^{-}}]={{10}^{-7}}M$ at ${{25}^{0}}C$
On diluting an acidic aqueous solution concentration of ${{H}^{+}}$ will decrease and pH of the solution will increase.

Complete Step by step solution:
Dilution means addition of solvent (water) in the solution, so that there is no change in the number of mole of solute or number of gram equivalent of solute remain unchanged before and after dilution.
Initially pH of the solution is 6. So ${{H}^{+}}$ concentration of the solution will be,
\[pH=-\log [{{H}^{+}}]\]
\[\Rightarrow -\log [{{H}^{+}}]=6\]
\[\Rightarrow [{{H}^{+}}]={{10}^{-6}}M\]
So according to the question on increasing dilution of the solution hundred times, so net concentration of the solute will be –
Initial concentration of the solution = ${{10}^{-6}}M$
After dilution solution concentration will be$\dfrac{{{10}^{-6}}}{100}={{10}^{-8}}M$
In ${{10}^{-8}}M$ acidic solution also contains water molecule there in which also undergo self-ionization
\[{{H}_{2}}O{{H}^{+}}+O{{H}^{-}}\] , $[{{H}^{+}}]={{10}^{-7}}M$ at ${{25}^{0}}C$
If it is taken simply here higher$[{{H}^{+}}]$ concentration of ${{H}_{2}}O$ molecule must be considered which is ${{10}^{-7}}M$ but$[{{H}^{+}}]$ from acid decreases the self-ionization of ${{H}_{2}}O$ molecule so, it will decrease $[{{H}^{+}}]$ from the water. Hence net concentration of $[{{H}^{+}}]$ must be smaller than ${{10}^{-7}}M$.
So net concentration of $[{{H}^{+}}]$ in the solution will be,
\[[{{H}^{+}}]={{10}^{-7}}+{{10}^{-8}}\]
\[\Rightarrow [{{H}^{+}}]={{10}^{-7}}\left[ 1+\dfrac{1}{10} \right]\]
By simplifying above equation, we get
$[{{H}^{+}}]=1.1\times {{10}^{-7}}M$
So final pH of the solution is,
\[pH=-\log [{{H}^{+}}]\]
\[\Rightarrow pH=-\log [1.1\times {{10}^{-7}}]\]
Thus, simplifying this, $pH=6.95$

So, option (A) will be the correct answer.

Note: For an acidic solution pH cannot be more than seven or seven. pH of a solution is unit less quantity. Dilution of acidic solution decreases the acidic nature of the solution. Change in pH of a solution occurs due to the change in the concentration of present free ${{H}^{+}}$ and $O{{H}^{-}}$ ion.