Chapter 8, Problem 29P

8-41 What is the [OH^-] and pOH of each solution?

(a) 0.10 M KOH,

(b) 0.10 M Na₂CO₃,

(c) 0.10 M Na₃PO₄,

(d) 0.10 M NaHCO₃,

Interpretation Introduction

(a)

Interpretation:

The hydroxide ion concentration and pOH of KOH to be identified.

Concept Introduction:

Dissociation of water results in production of hydronium and hydroxide ion. The dissociation constant of water is denoted by K_w and can be represented as follows:

[H₃ O+ ] [OH^- ] = k_w.

Where, k_w is dissociation constant of water = 10^-14.

[H₃ O+ ] is the concentration of hydronium ion.

[OH^- ] is the concentration of hydroxide ion.

Answer to Problem 29P

pOH of solution is 1.

Explanation of Solution

Hydronium ion concentration can be determined as:

$\begin{array}{l}\text{pH = - log }\left[{\text{H}}^{\text{+}}\right]\hfill \\ \text{13 = - log }\left[{\text{H}}^{\text{+}}\right]\hfill \\ \left[{\text{H}}^{\text{+}}\right]{\text{ = 10}}^{\text{-13}}\text{M}\hfill \end{array}$

Now, hydroxide ion concentration can be determined as:

$\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]{\text{ =10}}^{\text{-14}}\hfill \\ \left({\text{10}}^{\text{-13}}\right)\left[{\text{OH}}^{\text{-}}\right]{\text{ =10}}^{\text{-14}}\hfill \\ \left[{\text{OH}}^{\text{-}}\right]{\text{ =10}}^{\text{-1}}\text{M}\hfill \end{array}$

pOH can be determined as:

$\begin{array}{l}\text{pOH = -log }\left[{\text{OH}}^{\text{-}}\right]\hfill \\ {\text{= - log 10}}^{\text{-1}}\text{= 1}\hfill \end{array}$

Thus, pOH of solution is 1.

Interpretation Introduction

(b)

Interpretation:

The hydroxide ion concentration and pOH of Na₂ CO₃ to be identified.

Concept Introduction:

Dissociation of water results in production of hydronium and hydroxide ion. The dissociation constant of water is denoted by K_w and can be represented as follows:

[H₃ O+ ] [OH^- ] = k_w.

Where, k_w is dissociation constant of water = 10^-14.

[H₃ O+ ] is the concentration of hydronium ion.

[OH^- ] is the concentration of hydroxide ion.

Answer to Problem 29P

pOH of solution is 2.4.

Explanation of Solution

Hydronium ion concentration can be determined as:

$\begin{array}{l}\begin{array}{l}\text{pH = - log }\left[{\text{H}}^{\text{+}}\right]\hfill \\ \text{11}\text{.6= - log }\left[{\text{H}}^{\text{+}}\right]\hfill \end{array}\\ \left[{\text{H}}^{\text{+}}\right]\text{ = 2}{\text{.86 \&*\#x00A0;10}}^{\text{-12}}\text{M}\end{array}$

Now, hydroxide ion concentration can be determined as:

$\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]{\text{ =10}}^{\text{-14}}\hfill \\ \left(\text{2}{\text{.86 \&*\#x00A0;10}}^{\text{-12}}\right)\left[{\text{OH}}^{\text{-}}\right]{\text{ =10}}^{\text{-14}}\hfill \\ \left[{\text{OH}}^{\text{-}}\right]\text{ =3}{\text{.57 \&*\#x00A0;10}}^{\text{-3}}\text{M}\hfill \end{array}$

pOHcan be determined as:

$\begin{array}{l}\text{pOH = 14 – pH}\hfill \\ \text{ = 14 – 11}\text{.6 = 2}\text{.4}\hfill \end{array}$

Thus, pOH of solution is 2.4.

Interpretation Introduction

(c)

Interpretation:

The hydroxide ion concentration and pOH of Na₃ PO₄ to be identified.

Concept Introduction:

Dissociation of water results in production of hydronium and hydroxide ion. The dissociation constant of water is denoted by K_w and can be represented as follows:

[H₃ O+ ] [OH^- ] = k_w.

Where, k_w is dissociation constant of water = 10^-14.

[H₃ O+ ] is the concentration of hydronium ion.

[OH^- ] is the concentration of hydroxide ion.

Answer to Problem 29P

pOH of solution is 2.0.

Explanation of Solution

Hydronium ion concentration can be determined as:

$\begin{array}{l}\begin{array}{l}\text{pH = - log }\left[{\text{H}}^{\text{+}}\right]\hfill \\ \text{12= - log }\left[{\text{H}}^{\text{+}}\right]\hfill \end{array}\\ \left[{\text{H}}^{\text{+}}\right]{\text{ = 10}}^{\text{-12}}\text{M}\end{array}$

Now, hydroxide ion concentration can be determined as:

$\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]{\text{ =10}}^{\text{-14}}\hfill \\ \left({\text{10}}^{\text{-12}}\right)\left[{\text{OH}}^{\text{-}}\right]{\text{ =10}}^{\text{-14}}\hfill \\ \left[{\text{OH}}^{\text{-}}\right]{\text{ =10}}^{\text{-2}}\text{M}\hfill \end{array}$

pOHcan be determined as:

$\begin{array}{l}\text{pOH = 14 – pH}\hfill \\ \text{ = 14 – 12 = 2}\text{.0}\hfill \end{array}$

Thus, pOH of solution is 2.0.

Interpretation Introduction

(d)

Interpretation:

The hydroxide ion concentration and pOH of NaHCO₃ to be identified.

Concept Introduction:

Dissociation of water results in production of hydronium and hydroxide ion. The dissociation constant of water is denoted by K_w and can be represented as follows:

[H₃ O+ ] [OH^- ] = k_w.

Where, k_w is dissociation constant of water = 10^-14.

[H₃ O+ ] is the concentration of hydronium ion.

[OH^- ] is the concentration of hydroxide ion.

Answer to Problem 29P

pOH of solution is 5.6.

Explanation of Solution

Hydronium ion concentration can be determined as:

$\begin{array}{l}\text{pH = - log }\left[{\text{H}}^{\text{+}}\right]\hfill \\ \text{8}\text{.4= - log }\left[{\text{H}}^{\text{+}}\right]\hfill \\ \left[{\text{H}}^{\text{+}}\right]\text{ = 4}{\text{.12 \&*\#x00A0;10}}^{\text{-9}}\text{M}\hfill \end{array}$

Now, hydroxide ion concentration can be determined as:

$\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]{\text{ =10}}^{\text{-14}}\hfill \\ \left(\text{4}{\text{.12 \&*\#x00A0;10}}^{\text{-9}}\right)\left[{\text{OH}}^{\text{-}}\right]{\text{ =10}}^{\text{-14}}\hfill \\ \left[{\text{OH}}^{\text{-}}\right]\text{ =2}{\text{.43 \&*\#x00A0;10}}^{\text{-6}}\text{M}\hfill \end{array}$

pOH can be determined as:

$\begin{array}{l}\text{pOH = 14 – pH}\hfill \\ \text{ = 14 – 8}\text{.4 = 5}\text{.6}\hfill \end{array}$

Thus, pOH of solution is 5.6.