Question

Calculate the pH of $\text{ 1}{{\text{0}}^{-8}}\text{ M NaOH }$ the solution?

Answer 1

Hint: Every aqueous solution, whether acidic, alkaline, or neutral, contains both $\text{ }{{\text{H}}^{\text{+ }}}$ and $\text{ O}{{\text{H}}^{-}}\text{ }$ ions. Aqueous solution of sodium hydroxide is prepared in the water. Water itself dissociates into the $\text{ }{{\text{H}}^{\text{+ }}}$and $\text{ O}{{\text{H}}^{-}}\text{ }$ ions. Thus from the ionic product of water, the sum of $\text{ pOH }$ and $\text{ pH }$ is given as,
$\text{ p}{{\text{K}}_{\text{w}}}\text{=14 = pH + pOH }$
Since at room temperature pure water the concentrations of hydrogen ion and hydroxide ion is easy to $\text{ 1}{{\text{0}}^{-7}}\text{ }$.

Complete step by step answer:
Every aqueous solution, whether acidic, alkaline, or neutral, contains both $\text{ }{{\text{H}}^{\text{+ }}}$ and $\text{ O}{{\text{H}}^{-}}\text{ }$ ions. The product of the concentrations is always constant and equal to the $\text{ 1}\times \text{1}{{\text{0}}^{-14}}\text{ }$ at room temperature.
The $\text{ pH }$ is used to the hydrogen ion concentration like that hydroxyl ion concentration in the solution can be expressed as the $\text{ pOH }$ . The hydroxyl ion concentration is expressed as,
$\text{ pOH = }-\text{log}\left[ \text{O}{{\text{H}}^{-}} \right]\text{ }$
We have to determine the $\text{ pH }$ values of the $\text{ 1}{{\text{0}}^{-8}}\text{ M NaOH }$ . Sodium hydroxide dissociates as follows,
$\text{ NaOH }\to \text{ N}{{\text{a}}^{\text{+ }}}+\text{ O}{{\text{H}}^{-}}\text{ }$
The neutral water also contributes towards the hydroxide concentration. The water dissociates into hydrogen and hydroxyl ions. The equation is represented as follows,
$\text{ }{{\text{H}}_{\text{2}}}\text{O(}l)\text{ }\rightleftharpoons {{\text{H}}^{\text{+}}}\text{(}aq)+\text{O}{{\text{H}}^{-}}(aq)\text{ }$
Dissociation takes place to a very small extent .for pure and neutral water the concentration of hydrogen ion and the hydroxide ion formed is equal to the $\text{ 1}{{\text{0}}^{-7}}\text{ }$ .
$\text{ p}{{\text{K}}_{\text{w}}}\text{ = pH + pOH }$
The water provides $\text{ 1}{{\text{0}}^{-7}}\text{ M }$ hydroxide ion. Thus total hydroxide ion concentration in the solution is equal to hydroxide concentration from the water and hydroxide concentration from the sodium hydroxide solution. Thus total hydroxide in the solution is equal to,
$\text{ 1}{{\text{0}}^{-\text{7}}}\text{(from water) + 1}{{\text{0}}^{-\text{8}}}\text{(from NaOH) = 1}\text{.1}\times \text{1}{{\text{0}}^{-\text{7}}}\text{ M }$
Thus the $\text{ pH }$value of $\text{ NaOH }$the solution is,
$\begin{aligned} & \text{ pH =14}-\text{ }\left( -\log \left[ \text{O}{{\text{H}}^{-}} \right] \right) \\\ & \Rightarrow 14-\left( -\log (1.1\times {{10}^{-7}}) \right) \\\ & \therefore \text{pH = 14}-\text{6}\text{.96 = 7}\text{.04 } \\\ \end{aligned}$

Therefore, the $\text{ pH }$ value of the $\text{ 1}{{\text{0}}^{-8}}\text{ M NaOH }$ solution is $\text{ 7}\text{.04 }$.

Note: we know that the ionic product of the water is written as the product of hydrogen ion concentration and hydroxide ion concentration.$\text{ }{{\text{K}}_{\text{w}}}\text{ = }\left[ {{\text{H}}^{+}} \right]\left[ \text{O}{{\text{H}}^{-}} \right]\text{ }$ On taking the log and reversing we have, $\text{ p}{{\text{K}}_{\text{w}}}\text{ = pH + pOH }$.Thus without thinking we will use this formula and calculate the $\text{ pH }$ as 6 .however, this is a complete baseless as the solution becomes acidic on the addition of base. Thus it must be remembered that the water supplies the hydroxide ions to the solution too.