Chapter 15, Problem 15.28P

Interpretation Introduction

Interpretation:

The pH of the unknown $\text{+50}\text{.0}\text{mV}$ and slope of the calibration curve $\left(\text{mV/pH}\text{unit}\right)$ and the theoretical slope at ${\text{20}}^{\text{0}}\text{C}$ and the value of $\text{β}$ has to be calculated.

Concept Introduction:

Calibrating a glass electrode:

When u calibrate an electrode with standard buffers, to calculate the voltage with each buffer. The pH of buffer $\text{S1}$ is ${\text{pH}}_{\text{S1}}$ and measured electrode potential in that buffer is ${\text{E}}_{\text{S1}}$.

The pH of buffer $\text{S2}$ is ${\text{pH}}_{\text{S2}}$ and measured electrode potential in that buffer is ${\text{E}}_{\text{S2}}$.

$\frac{{\text{E-E}}_{\text{S1}}}{{\text{pH-pH}}_{\text{S1}}}\text{=}\frac{{\text{E}}_{\text{S2}}{\text{-E}}_{\text{S1}}}{{\text{pH}}_{\text{S2}}{\text{-pH}}_{\text{S1}}}$

The slope of the line is

$\frac{\text{ΔE}}{\text{ΔpH}}\text{=}\frac{{\text{E}}_{\text{S2}}{\text{-E}}_{\text{S1}}}{{\text{pH}}_{\text{S2}}{\text{-pH}}_{\text{S1}}}$

Theoretical slope= $\frac{\text{RTln10}}{\text{F}}$

$\text{β}\text{=}\frac{\text{observed}\text{slope}}{\text{theoretical}\text{slope}}$