Question

$ {10^{ - 7}}M $ Question: pH of solution of strong acid is $ 4 $ . What will be the pH of the solution obtained after diluting the given acid solution $ 1000 $ times?
(A) $ 6.7 $
(B) $ 7 $
(C) $ 6.9 $
(D) $ 7.3 $

Answer 1

Hint: We know that the pH value of any substance is dependent upon the concentration of $ H^+ $ and $ OH^- $ ions in the aqueous solution. Since we need to find out the pH of the solution after dilution, we will consider the relation of pH and the concentration of $ H^+ $ ions.

Complete answer:
Here we are given that the pH of the strong acid is $ 4 $ , so we know the formula to find out the pH of any aqueous solution, that is:
 $ pH = - log [H^+ ] $ $ - - - (1) $
Since we the know the pH value of the strong acid before dilution so we will put the value in the relation, we get
 $ 4 = - log [H^+ ] $
After solving we get,
 $ [H^+ ] = {10^{ - 4}}M $
Now, we will solve for the $ [H^+ ] $ after $ 1000 $ times dilution,
 $ [H^+ ] = \dfrac{{{{10}^{ - 4}}}}{{{{10}^3}}} = {10^{ - 7}}M $
Under the dilution condition, water molecule will dissociate itself into hydrogen and hydroxide ion:
 $ {H_2}O \rightleftharpoons {H^ + } + O{H^ - } $
Now, we will find out the total concentration of $ {H^ + } $ ions, since we know that $ [H^+ ] = {10^{ - 7}} $ for $ {H_2}O $ molecule
Therefore, Total concentration of $ H^+ = {10^{ - 7}} + {10^{ - 7}} = 2 \times {10^{ - 7}} $ M
Now, we have the total concentration of Hydrogen ion, so we will put this value in the equation $ (1) $ :
 $ pH = - log [{H^ + }] = - log (2 \times {10^{ - 7}}) $
 $ = 7 - log 2 $ , value of log 2 is $ 0.301 $
 $ = 7 - 0.301 = 6.699 \sim 6.7 $
This is our final value of pH after the dilution of strong acid.
Therefore, the correct option is (A) $ 6.7 $ .

Note:
When we solve such types of questions, it looks easy to solve but we need to keep the concentration of the hydrogen ions which are provided by water molecules in mind, which is $ {10^{ - 7}}M $ .