Chapter 10, Problem 10.1P

Interpretation Introduction

Interpretation:

Chemistry that goes into each region of titration between ${\mathrm{OH}}^{-}$ and $H^{+}$ ions has to be explained.

Concept Introduction:

In an acid base-kind of titration, progress of reaction is quantified by neutralization of analyte when definite volume of titrant is added. The indicators used are sensitive to $\text{pH}$ changes. Suitability of such indicator is governed by their ability to detect end point accurately when the equivalence point is reached.

Explanation of Solution

Before the equivalence point is reached any volume of $H^{+}$ consumes the corresponding stoichiometric volume of ${\mathrm{OH}}^{-}$ in accordance with neutralization principle. However, concentration of ${\mathrm{OH}}^{-}$ ion remains in excess at any point before titration. Hence $\text{pH}$ at this point is more than 7. Manner $\text{pH}$ is calculated as follows:

  ${\left[{\mathrm{OH}}^{-}\right]}_{\text{left}}={\left[{\mathrm{OH}}^{-}\right]}_{\text{initial}}-{\left[{\mathrm{OH}}^{-}\right]}_{\text{consumed}}$

Net volume is calculated as follows:

  $\text{Net volume}=\text{Volume of analyte in flask}+\text{Volume added}$

New concentration of $\left[{\mathrm{OH}}^{-}\right]$ is calculated as follows:

  $\left[{\mathrm{OH}}^{-}\right]=\frac{{\left[{\mathrm{OH}}^{-}\right]}_{\text{left}}}{\text{Net volume}}$

Expression to calculate $\left[H^{+}\right]$ from ionic product of water is as follows:

  $\left[H^{+}\right]=\frac{K_w}{\left[{\text{OH}}^{-}\right]}$

With $\left[H^{+}\right]$ thus obtained  $\text{pH}$ is determined by expression as follows:

  $\text{pH}=-\text{log}\left[H^{+}\right]$

Once equivalence is reached entire ${\mathrm{OH}}^{-}$ ion has been neutralized by sufficient $H^{+}$ ion. Thus there is no excess of ${\mathrm{OH}}^{-}$ and thus $\text{pH}$ at this point is 7. At this point, there is equal concentration of ${\mathrm{OH}}^{-}$ and $H^{+}$ ion.

Beyond equivalence point, all added $H^{+}$ will lead to excess concentration of $H^{+}$ ion and hence $\text{pH}$ at this point is less than 7. Here excess $H^{+}$ is determined as follows:

  $\text{Excess }{H}^{+}=\left(\text{Volume}\right)\left(\text{Molarity}\right)$

Net volume is calculated as follows:

  $\text{Net volume}=\text{Volume of analyte in flask}+\text{Volume added}$

New concentration of $\left[H^{+}\right]$ is calculated as follows:

  $\left[H^{+}\right]=\frac{\text{Excess }{H}^{+}\left(\text{mmol}\right)}{\text{Net volume}\left(\text{mL}\right)}$

With $\left[H^{+}\right]$ thus obtained, $\text{pH}$ is again determined by expression as follows:

  $\text{pH}=-\text{log}\left[H^{+}\right]$