Question

The pH of 10-8 M HCl solution is:
a.) 8
b.) 1
c.) Close to 7
d.) 0

Answer 1

Hint: pH of a solution is equal to the negative logarithm of the concentration of the protons or more precisely the activity of the protons. When the solution is dilute, we can use the concentration instead of the activity.

Complete step by step solution:
pH is a scale that is used to denote the acidity or basicity of an aqueous solution. Lower the pH value, more acidic the solution is; higher the pH value, more basic the solution is. At room temperature (25°C) pure water is neutral and has a pH of 7.
The pH is the negative logarithm of the concentration of hydrogen ions present in a solution. For denoting the concentration of a solution, we use Molarity.
At 25 °C, if the pH of a solution is less than 7, it is acidic while if it is greater than 7 then the solution is basic. The range of pH scale at 25 °C when the solvent is water is between 0-14 inclusive of the values denoting the range. For the above question, the concentration of HCl is given as 10−8 M. From this it seems that the pH should be 8, but this is a very dilute solution of HCl therefore we cannot neglect the self-ionization of water in this case.
Therefore, $ Total\quad \left[ H^{ + } \right] =\left[ { H }^{ + } \right] \quad from\quad HCl+\left[ { H }^{ + } \right] \quad from\quad water $
Since HCl is a strong acid, it completely dissociates.
So, $\quad \left[ { H }^{ + } \right] \quad from\quad HCl={ 10 }^{ -8 }M$
Since water dissociates into protons and hydroxide ions partially, therefore HCl will make the ionization of water to move towards the left, therefore the concentration of protons from water will be less than $ { 10 }^{ -7 }M $. Let this concentration be equal to x.
$ \left[ { H }^{ + } \right] \quad from\quad water=\left[ { OH }^{ - } \right] from\quad water=x $
$Total\quad \left[ { H }^{ + } \right] =x+{ 10 }^{ -8 }M\quad$.......(1)
But at 25 °C,
$ \left[ { H }^{ + } \right] \left[ { OH }^{ - } \right] =1.0\times { 10 }^{ -14 }(ionic\quad product\quad of\quad water) $
=$ ({ 10 }^{ -8 }+x)\times x=1.0\times { 10 }^{ -14 } $
=$ { x }^{ 2 }+{ 10 }^{ -8 }x-{ 10 }^{ -14 }=0 $
By solving this quadratic equation, we get
$ { x }=9.5\times { 10 }^{ -8 } $
Substituting the above value in equation (1), we get
$ Total\quad \left[ { H }^{ + } \right] =(9.5\times { 10 }^{ -8 })+{ 10 }^{ -8 }\quad M$
Therefore, $ Total\quad \left[ { H }^{ + } \right] =1.05\times { 10 }^{ -7 } M$.
Now,
$ pH=-\log { \left[ { H }^{ + } \right] }$
$ pH=-\log { \left[ 1.05\times { 10 }^{ -7 } \right] } $
$ pH=6.98 $

So, the correct answer is “Option C”.

Note: The pH range depends upon the type of solvent along with the temperature conditions. If the solvent is water and the temperature is greater than 25 °C, then the neutral pH will be less than 7.