Chapter 15, Problem 15.82SP

(a)

Interpretation Introduction

Interpretation:

$\text{pH}$ of the solution should be calculated, after addition of $\text{10}\text{.0}\text{mL}\text{of}\text{0}\text{.100 M}\text{NaOH}$ in $\text{50}\text{.0}\text{mL}\text{of }\text{0}\text{.100}\text{M}$ alanine solution.

Concept Introduction:

$\text{pH}$ of the buffer solution is given by Henderson-Hasselbach Equation and it is,

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[Base]}}{\text{[Acid]}}$

(b)

Interpretation Introduction

Interpretation:

$\text{pH}$ of the solution should be calculated, after addition of $\text{25}\text{.0}\text{mL}\text{of}\text{0}\text{.100 M}\text{NaOH}$ in $\text{50}\text{.0}\text{mL}\text{of }\text{0}\text{.100}\text{M}$ alanine solution.

Concept Introduction:

$\text{pH}$ of the buffer solution is given by Henderson-Hasselbach Equation and it is,

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[Base]}}{\text{[Acid]}}$

(c)

Interpretation Introduction

Interpretation:

$\text{pH}$ of the solution should be calculated, after addition of $\text{50}\text{.0}\text{mL}\text{of}\text{0}\text{.100 M}\text{NaOH}$ in $\text{50}\text{.0}\text{mL}\text{of }\text{0}\text{.100}\text{M}$ alanine solution.

Concept Introduction:

$\text{pH}$ of the buffer solution is given by Henderson-Hasselbach Equation and it is,

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[Base]}}{\text{[Acid]}}$

(d)

Interpretation Introduction

Interpretation:

$\text{pH}$ of the solution should be calculated, after addition of $\text{75}\text{.0}\text{mL}\text{of}\text{0}\text{.100 M}\text{NaOH}$ in $\text{50}\text{.0}\text{mL}\text{of }\text{0}\text{.100}\text{M}$ alanine solution.

Concept Introduction:

$\text{pH}$ of the buffer solution is given by Henderson-Hasselbach Equation and it is,

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[Base]}}{\text{[Acid]}}$

(e)

Interpretation Introduction

Interpretation:

$\text{pH}$ of the solution should be calculated, after addition of $\text{100}\text{.0}\text{mL}\text{of}\text{0}\text{.100 M}\text{NaOH}$ in $\text{50}\text{.0}\text{mL}\text{of }\text{0}\text{.100}\text{M}$ alanine solution.

Concept Introduction:

$\text{pH}$:

The negative log of base 10 ${\text{H}}^{+}$ ion concentration is given by $\text{pH}$ of the solution.

$\text{pH}\text{=}{\text{-log}}_{\text{10}}{\text{[H}}^{\text{+}}\text{]}$

Equilibrium constant for base dissociation in aqueous solution is,

The general base dissociation reaction in water is,

$B+H_{2}O\stackrel{}{\rightleftharpoons }H{B}^{+}+O{H}^{-}$

$\text{Equilibrium}\text{constant}{\text{K}}_b\text{=}\frac{\left[B{H}^{+}\right]\left[O{H}^{-}\right]}{\left[B\right]}$