Question

Calculate the $\text{ pH }$ of $\text{ 1}{{\text{0}}^{-8}}\text{M HCl }$.
A) 8
B) 6
C) 7
D) $\text{ 6}\text{.98 }$

Answer 1

Hint: The hydrogen ion concentration is calculated in terms of $\text{ pH }$ . The $\text{ pH }$is a negative logarithmic value of a hydrogen concentration. It is written as:
$\text{ pH = }-\log \left[ {{\text{H}}^{+}} \right]\text{ }$
However, the direct implication of the formula is not advisable. The concentration of hydrochloric acid is very dilute thus water cannot be neglected.

Complete Solution :
Firstly, the concentration of $\text{ HCl }$ is given as, $\text{ }\left[ \text{ HCl} \right]\text{ = 1}{{\text{0}}^{-8}}\text{M }$
It is a very dilute acidic solution when $\text{ }{{\text{H}}^{+}}\text{ }$concentrations from acid, and water are comparable, the concentration of $\text{ }{{\text{H}}^{+}}\text{ }$obtained from water cannot be neglected.
Thus, the total hydrogen ion concentration in the solution is obtained from the $\text{ HCl }$ and the $\text{ }{{\text{H}}^{+}}\text{ }$obtained from the water.
$\text{ Total }\left[ {{\text{H}}^{\text{+}}} \right]\text{ = }{{\left[ {{\text{H}}^{\text{+}}} \right]}_{\text{HCl}}}\text{ + }{{\left[ {{\text{H}}^{\text{+}}} \right]}_{{{\text{H}}_{\text{2}}}\text{O}}}\text{ }$
Here, $\text{ HCl }$ is a strong acid, so it completely ionizes. The ionization of hydrochloric acid is as follows,
$\text{ HCl }\to \text{ }{{\text{H}}^{\text{+}}}\text{ + C}{{\text{l}}^{-}}\text{ }$
Thus, $\text{ }{{\left[ {{\text{H}}^{\text{+}}} \right]}_{\text{HCl}}}\text{ = 1}\text{.8}\times \text{1}{{\text{0}}^{-8}}\text{ M }$

- Now, the concentration of hydrogen ions from the water is calculated from the ionization of water. From ionization, the concentration of hydrogen ions is equal to the $\text{ }\left[ \text{O}{{\text{H}}^{-}} \right]\text{ }$ ions let’s consider the concentration of \[{{\text{H}}^{+}}\] and $\text{O}{{\text{H}}^{-}}$ as x. The ionization of water would be,
$\text{ }\begin{matrix}
   {} & {{\text{H}}_{\text{2}}}\text{O} & \rightleftharpoons & {{\text{H}}^{+}} & + & \text{O}{{\text{H}}^{-}} \\
   \text{without acid} & {} & {} & x & {} & x \\
   \text{With acid} & {} & {} & x+(1.0\times {{10}^{-8}}) & {} & x \\
\end{matrix}$

- In absence of acid, the ionization of water is written as:
$\text{ }{{\left[ {{\text{H}}^{\text{+}}} \right]}_{{{\text{H}}_{\text{2}}}\text{O}}}\text{ = }{{\left[ \text{O}{{\text{H}}^{-}} \right]}_{{{\text{H}}_{\text{2}}}\text{O}}}\text{ = x }$
In presence of an acid, the concentration of hydrogen ion and hydroxide ion is written as,
$\text{ }{{\left[ {{\text{H}}^{\text{+}}} \right]}_{{{\text{H}}_{\text{2}}}\text{O + HCl}}}\text{=x + (1}\text{.0 }\times \text{1}{{\text{0}}^{-8}}\text{) , }{{\left[ \text{O}{{\text{H}}^{-}} \right]}_{{{\text{H}}_{\text{2}}}\text{O}}}\text{ = x }$

- We know that the concentration product of hydrogen ion and hydroxide ion is:
$\text{ }\left[ {{\text{H}}^{\text{+}}} \right]\text{ }\left[ \text{O}{{\text{H}}^{-}} \right]\text{ = 1}{{\text{0}}^{-14}}\text{ }$

- Now, substituting the values:
$\begin{align}
  & \text{ 1}{{\text{0}}^{-14}}=\left[ {{\text{H}}^{\text{+}}} \right]\text{ }\left[ \text{O}{{\text{H}}^{-}} \right]\text{ } \\
 & \Rightarrow \text{1}{{\text{0}}^{-14}}=(\text{x + 1}\text{.0 }\times \text{1}{{\text{0}}^{-8}}\text{)(x)} \\
 & \Rightarrow \text{x = 9}\text{.5 }\times \text{1}{{\text{0}}^{-8}}\text{ } \\
\end{align}$

Thus, the total concentration of hydrogen ion in the solution is:
$\text{ Total }\left[ {{\text{H}}^{\text{+}}} \right]\text{ = 1}\text{.8}\times \text{1}{{\text{0}}^{-8}}\text{ M + 9}\text{.5}\times \text{1}{{\text{0}}^{-8}}\text{ M = 1}\text{.05 }\times \text{1}{{\text{0}}^{-7}}\text{ M }$

- Now let's find out the $\text{ pH }$ of the hydrogen ion.
$\begin{align}
 & \text{ pH = }-\log \left[ {{\text{H}}^{+}} \right] \\
 & \Rightarrow -\log \left( 1.05\times {{10}^{-7}} \right) \\
 & \therefore \text{pH = 6}\text{.978 }\simeq \text{ 6}\text{.98 } \\
\end{align}$
So, the correct answer is “Option D”.

Note: There is a relation to find the pH, i.e.$\text{ pH = }-\log \left[ {{\text{H}}^{+}} \right]\text{ }$ , but in this case, we get pH equal to 8. For an acidic solution, the value of pH can’t be greater than 7. Thus, here we cannot use the direct formula instead of that we consider the hydrogen ion from the water. The diluted solution would be in the desired range.