Chapter 15, Problem 15.29P

a)

Interpretation Introduction

Interpretation:

The ionic strength of the buffer is $\text{μ=0}\text{.100}\text{m}$ has to be proved.

Explanation of Solution

Given:

pH is $\text{6}\text{.865}$ at ${\text{25}}^{\text{0}}\text{C}$ for $\text{0}{\text{.0250mKH}}_{\text{2}}{\text{PO}}_{\text{4}}|\text{0}{\text{.0250mNa}}_{\text{2}}{\text{HPO}}_{\text{4}}$ buffer.

The change in buffer species when mixed with the acid ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}$ with its conjugate base, ${\text{HPO}}_{\text{4}}^{\text{2-}}$ are negligible. The ionic strength of $\text{0}\text{.0250}\text{m}{\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}$ $\left(\text{1:1}\text{electrolyte}\right)$ is $\text{0}\text{.0250}\text{m}$ and the ionic strength of $\text{0}\text{.0250}\text{m}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ $\left(\text{2:1}\text{eletrolyte}\right)$ is $\text{0}\text{.0750}\text{m}$ so the total ionic strength is $\text{0}\text{.100}\text{m}$.

Conclusion

The ionic strength of the buffer is $\text{μ=0}\text{.100}\text{m}$ was proved.

b)

Interpretation Introduction

Interpretation:

The pH and ${\text{K}}_{\text{2}}$ for phosphoric acid, the quotient of activity coefficients ${\text{γHPO}}_{\text{4}}^{\text{-}}{\text{/γH}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\text{at}\text{μ=0}\text{.100}\text{m}$ has to be calculated.

Concept Introduction:

Metal ion buffer:

A metal-ion buffer gives a controlled source of free metal ions in a way related to the regulation of hydrogen ion concentration by a pH buffer. A metal-ion buffer solution has the free (hydrated) metal ion along with a complex compound made by the association of the ion with a ligand in excess. The concentration of free metal ion mainly depends on the total concentration of each component (ligand and metal ion) and also on the stability constant of the complex. The concentration of the free metal ion depends upon on solution pH if the ligand can undergo protonation.

Concentration= $\frac{\left[\text{M}\right]\left[\text{L}\right]}{\left[\text{ML}\right]}$

Answer to Problem 15.29P

The pH and ${\text{K}}_{\text{2}}$ for phosphoric acid, the quotient of activity coefficients ${\text{γHPO}}_{\text{4}}^{\text{-}}{\text{/γH}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\text{at}\text{μ=0}\text{.100}\text{m}$ was calculated as ${\text{K}}_{\text{2}}\text{=0}\text{.4645}$ .

Explanation of Solution

Given:

Concentration= $\frac{\left[\text{M}\right]\left[\text{L}\right]}{\left[\text{ML}\right]}$

${\text{K}}_{\text{2}}\text{=}\frac{\left[{\text{H}}^{\text{+}}\right]{\text{γH}}^{\text{+}}\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]{\text{γHPO}}_{\text{4}}^{\text{2-}}}{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]{\text{γH}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}}$

Given ${\text{K}}_{\text{2}}{\text{=10}}^{\text{-7}\text{.198}}$ and $\left[{\text{H}}^{\text{+}}\right]{\text{γHI}}^{\text{+}}{\text{=10}}^{\text{-pH}}$

Where pH= $\text{6}\text{.685}$

$\left[{\text{H}}^{\text{+}}\right]{\text{γHI}}^{\text{+}}{\text{=10}}^{\text{-pH}}{\text{=10}}^{\text{-6}\text{.865}}$

Then,

$\frac{{\text{γHPO}}_{\text{4}}^{\text{2-}}}{{\text{γH}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{2-}}}\text{=}\frac{{\text{K}}_{\text{2}}\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]}{\left[{\text{H}}^{\text{+}}\right]{\text{γH}}^{\text{+}}\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]}\text{=}\frac{{\text{10}}^{\text{-7}\text{.168}}\left[\text{0}\text{.0250}\right]}{{\text{10}}^{\text{-6}\text{.685}}\left[\text{0}\text{.0250}\right]}\text{=0}\text{.4645}$

Units will cancel each other since, both the numerator and denominator has concentrations.

Conclusion

The pH and ${\text{K}}_{\text{2}}$ for phosphoric acid, the quotient of activity coefficients ${\text{γHPO}}_{\text{4}}^{\text{-}}{\text{/γH}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\text{at}\text{μ=0}\text{.100}\text{m}$ was calculated as ${\text{K}}_{\text{2}}\text{=0}\text{.4645}$ .

c)

Interpretation Introduction

Interpretation:

The needed molalities of ${\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}\text{and}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ should be mixed to obtain a pH of $\text{7}\text{.000}$ and $\text{μ=0}\text{.100}\text{m}$ has to be explained.

Concept Introduction:

Metal ion buffer:

A metal-ion buffer gives a controlled source of free metal ions in a way related to the regulation of hydrogen ion concentration by a pH buffer. A metal-ion buffer solution has the free (hydrated) metal ion along with a complex compound made by the association of the ion with a ligand in excess. The concentration of free metal ion mainly depends on the total concentration of each component (ligand and metal ion) and also on the stability constant of the complex. The concentration of the free metal ion depends upon on solution pH if the ligand can undergo protonation.

Concentration= $\frac{\left[\text{M}\right]\left[\text{L}\right]}{\left[\text{ML}\right]}$

Answer to Problem 15.29P

The needed molalities of ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\text{and}{\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}$ should be mixed to obtain a pH of $\text{7}\text{.000}$ and $\text{μ=0}\text{.100}\text{m}$ was $\text{0}\text{.0268}\text{m,}\text{0}\text{.0196}\text{m}$ .

Explanation of Solution

The pH $\text{7}\text{.000}$ can be obtained by decreasing the concentration of acid $\left({\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right)$ and increase the concentration of base $\left({\text{HPO}}_{\text{4}}^{\text{-}}\right)$.

In order maintain a constant ionic strength we must decrease ${\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}$ three times as much as we increase ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ because ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ can contributes three times as much as ${\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}$ to the ionic strength.

So increasing the ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ by $\text{x}$ and decrease ${\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}$ by $\text{3x}$.

Given:

Concentration= $\frac{\left[\text{M}\right]\left[\text{L}\right]}{\left[\text{ML}\right]}$

${\text{K}}_{\text{2}}\text{=}\frac{\left[{\text{H}}^{\text{+}}\right]{\text{γH}}^{\text{+}}\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]{\text{γHPO}}_{\text{4}}^{\text{2-}}}{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]{\text{γH}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}}$

Given ${\text{K}}_{\text{2}}{\text{=10}}^{\text{-7}\text{.198}}$

${\text{10}}^{\text{-7}\text{.198}}\text{=}\frac{{\text{10}}^{\text{-7}\text{.000}}\left[\text{0}\text{.0250+x}\right]}{\left[\text{0}\text{.0250-3x}\right]}\left(\text{0}\text{.4645}\right)$

$\text{x=0}\text{.0018}\text{m}$

So, the new concentration are ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\text{=0}\text{.0268}\text{m}$ and ${\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}\text{=0}\text{.0196}\text{m}$

Conclusion

The needed molalities of ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\text{and}{\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}$ should be mixed to obtain a pH of $\text{7}\text{.000}$ and $\text{μ=0}\text{.100}\text{m}$ was $\text{0}\text{.0268}\text{m,}\text{0}\text{.0196}\text{m}$ .