Chapter 6, Problem 6B.1BST

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{0}\text{.077}\text{M}\text{NaOH(aq)}$ has to be calculated.

Concept Introduction:

Autopyrolysis is a reaction in which proton is transfer from one molecule to another molecule of same kind.  Autopyrolysis constant can be expressed by the equation shown below;

    $K_W=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]$

For pure water the value of autopyrolysis constant is equal to $1.0\times {10}^{-14}$ at $\text{25°C}$.  Unit of autopyrolysis constant is ${\text{mol}}^{\text{2}}\cdot {\text{L}}^{-2}$.  This is because, the unit for concentration of hydroxide ion and hydronium ion is $\text{mol}\cdot {\text{L}}^{-1}$.

Negative logarithm of the concentration of hydronium ion is the $\text{pH}$ of solution.  This can be given in equation form as follows;

    $\text{pH}=-\log [{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}]$