Chapter 9, Problem 9.28E

Calculate the molar concentration of ${\text{OH}}^{-}$ in water solutions with the following ${\text{H}}_{\text{3}}{\text{O}}^{+}$ molar concentrations:

a. $0.044$

b. $1.3\times {10}^{-4}$

c. $0.0087$

d. $7.9\times {10}^{-10}$

e. $3.3\times {10}^{-2}$

Interpretation Introduction

(a)

Interpretation:

The molar concentration of ${\text{OH}}^{-}$ in water solutions with the given ${\text{H}}_{\text{3}}{\text{O}}^{+}$ molar concentrations is to be calculated.

Concept introduction:

The water undergoes self-ionization which can be represented by the reaction,

${\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)⥄{\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)$

The ionization constant of water is represented as,

$K=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_{\text{2}}\text{O}\right]\left[{\text{H}}_{\text{2}}\text{O}\right]}$

The concentration of water remains constant and the self-ionization constant of water becomes,

$\begin{array}{l}{K}_{\text{w}}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]\\ K_{\text{w}}=\left(1.0\times {10}^{-7}\text{mol}/\text{L}\right)\left(1.0\times {10}^{-7}\text{mol}/\text{L}\right)\\ K_{\text{w}}=1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2\end{array}$

Answer to Problem 9.28E

The molar concentration of ${\text{OH}}^{-}$ in water solutions is $2.27\times {10}^{-13}\text{mol}/\text{L}$.

Explanation of Solution

The ionic product of water $K_{\text{w}}$ is,

$K_{\text{w}}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$

The value of $K_{\text{w}}$ is $1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2$.

The given ${\text{H}}_{\text{3}}{\text{O}}^{+}$ molar concentration is $0.044\text{mol}/\text{L}$. Substitute this value in the formula for ionic product.

$\begin{array}{rll}1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2& =& 0.044\text{mol}/\text{L}\left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]& =& \frac{1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2}{0.044\text{mol}/\text{L}}\\ \left[{\text{OH}}^{-}\right]& =& 2.27\times {10}^{-13}\text{mol}/\text{L}\end{array}$

Thus, the molar concentration of ${\text{OH}}^{-}$ in water solutions is $2.27\times {10}^{-13}\text{mol}/\text{L}$.

Conclusion

The molar concentration of ${\text{OH}}^{-}$ in water solutions is $2.27\times {10}^{-13}\text{mol}/\text{L}$.

Interpretation Introduction

(b)

Interpretation:

The molar concentration of ${\text{OH}}^{-}$ in water solutions with the given ${\text{H}}_{\text{3}}{\text{O}}^{+}$ molar concentrations is to be calculated.

Concept introduction:

The water undergoes self-ionization which can be represented by the reaction,

${\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)⥄{\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)$

The ionization constant of water is represented as,

$K=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_{\text{2}}\text{O}\right]\left[{\text{H}}_{\text{2}}\text{O}\right]}$

The concentration of water remains constant and the self-ionization constant of water becomes,

$\begin{array}{l}{K}_{\text{w}}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]\\ K_{\text{w}}=\left(1.0\times {10}^{-7}\text{mol}/\text{L}\right)\left(1.0\times {10}^{-7}\text{mol}/\text{L}\right)\\ K_{\text{w}}=1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2\end{array}$

Answer to Problem 9.28E

The molar concentration of ${\text{OH}}^{-}$ in water solutions is $7.69\times {10}^{-11}\text{mol}/\text{L}$.

Explanation of Solution

The ionic product of water $K_{\text{w}}$ is,

$K_{\text{w}}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$

The value of $K_{\text{w}}$ is $1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2$.

The given ${\text{H}}_{\text{3}}{\text{O}}^{+}$ molar concentration is $1.3\times {10}^{-4}\text{mol}/\text{L}$. Substitute this value in the formula for ionic product.

$\begin{array}{rll}1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2& =& 1.3\times {10}^{-4}\text{mol}/\text{L}\left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]& =& \frac{1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2}{1.3\times {10}^{-4}\text{mol}/\text{L}}\\ \left[{\text{OH}}^{-}\right]& =& 7.69\times {10}^{-11}\text{mol}/\text{L}\end{array}$

Thus, the molar concentration of ${\text{OH}}^{-}$ in water solutions is $7.69\times {10}^{-11}\text{mol}/\text{L}$.

Conclusion

The molar concentration of ${\text{OH}}^{-}$ in water solutions is $7.69\times {10}^{-11}\text{mol}/\text{L}$.

Interpretation Introduction

(c)

Interpretation:

The molar concentration of ${\text{OH}}^{-}$ in water solutions with the given ${\text{H}}_{\text{3}}{\text{O}}^{+}$ molar concentrations is to be calculated.

Concept introduction:

The water undergoes self-ionization which can be represented by the reaction,

${\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)⥄{\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)$

The ionization constant of water is represented as,

$K=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_{\text{2}}\text{O}\right]\left[{\text{H}}_{\text{2}}\text{O}\right]}$

The concentration of water remains constant and the self-ionization constant of water becomes,

$\begin{array}{l}{K}_{\text{w}}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]\\ K_{\text{w}}=\left(1.0\times {10}^{-7}\text{mol}/\text{L}\right)\left(1.0\times {10}^{-7}\text{mol}/\text{L}\right)\\ K_{\text{w}}=1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2\end{array}$

Answer to Problem 9.28E

The molar concentration of ${\text{OH}}^{-}$ in water solutions is $1.15\times {10}^{-12}\text{mol}/\text{L}$.

Explanation of Solution

The ionic product of water $K_{\text{w}}$ is,

$K_{\text{w}}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$

The value of $K_{\text{w}}$ is $1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2$.

The given ${\text{H}}_{\text{3}}{\text{O}}^{+}$ molar concentration is $0.0087\text{mol}/\text{L}$. Substitute this value in the formula for ionic product.

$\begin{array}{rll}1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2& =& 0.0087\text{mol}/\text{L}\left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]& =& \frac{1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2}{0.0087\text{mol}/\text{L}}\\ \left[{\text{OH}}^{-}\right]& =& 1.15\times {10}^{-12}\text{mol}/\text{L}\end{array}$

Thus, the molar concentration of ${\text{OH}}^{-}$ in water solutions is $1.15\times {10}^{-12}\text{mol}/\text{L}$.

Conclusion

The molar concentration of ${\text{OH}}^{-}$ in water solutions is $1.15\times {10}^{-12}\text{mol}/\text{L}$.

Interpretation Introduction

(d)

Interpretation:

The molar concentration of ${\text{OH}}^{-}$ in water solutions with the given ${\text{H}}_{\text{3}}{\text{O}}^{+}$ molar concentrations is to be calculated.

Concept introduction:

The water undergoes self-ionization which can be represented by the reaction,

${\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)⥄{\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)$

The ionization constant of water is represented as,

$K=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_{\text{2}}\text{O}\right]\left[{\text{H}}_{\text{2}}\text{O}\right]}$

The concentration of water remains constant and the self-ionization constant of water becomes,

$\begin{array}{l}{K}_{\text{w}}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]\\ K_{\text{w}}=\left(1.0\times {10}^{-7}\text{mol}/\text{L}\right)\left(1.0\times {10}^{-7}\text{mol}/\text{L}\right)\\ K_{\text{w}}=1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2\end{array}$

Answer to Problem 9.28E

The molar concentration of ${\text{OH}}^{-}$ in water solutions is $1.26\times {10}^{-5}\text{mol}/\text{L}$.

Explanation of Solution

The ionic product of water $K_{\text{w}}$ is,

$K_{\text{w}}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$

The value of $K_{\text{w}}$ is $1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2$.

The given ${\text{H}}_{\text{3}}{\text{O}}^{+}$ molar concentration is $7.9\times {10}^{-10}\text{mol}/\text{L}$. Substitute this value in the formula for ionic product.

$\begin{array}{rll}1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2& =& 7.9\times {10}^{-10}\text{mol}/\text{L}\left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]& =& \frac{1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2}{7.9\times {10}^{-10}\text{mol}/\text{L}}\\ \left[{\text{OH}}^{-}\right]& =& 1.26\times {10}^{-5}\text{mol}/\text{L}\end{array}$

Thus, the molar concentration of ${\text{OH}}^{-}$ in water solutions is $1.26\times {10}^{-5}\text{mol}/\text{L}$.

Conclusion

The molar concentration of ${\text{OH}}^{-}$ in water solutions is $1.26\times {10}^{-5}\text{mol}/\text{L}$.

Interpretation Introduction

(e)

Interpretation:

The molar concentration of ${\text{OH}}^{-}$ in water solutions with the given ${\text{H}}_{\text{3}}{\text{O}}^{+}$ molar concentrations is to be calculated.

Concept introduction:

The water undergoes self-ionization which can be represented by the reaction,

${\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)⥄{\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)$

The ionization constant of water is represented as,

$K=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_{\text{2}}\text{O}\right]\left[{\text{H}}_{\text{2}}\text{O}\right]}$

The concentration of water remains constant and the self-ionization constant of water becomes,

$\begin{array}{l}{K}_{\text{w}}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]\\ K_{\text{w}}=\left(1.0\times {10}^{-7}\text{mol}/\text{L}\right)\left(1.0\times {10}^{-7}\text{mol}/\text{L}\right)\\ K_{\text{w}}=1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2\end{array}$

Answer to Problem 9.28E

The molar concentration of ${\text{OH}}^{-}$ in water solutions is $3.03\times {10}^{-13}\text{mol}/\text{L}$.

Explanation of Solution

The ionic product of water $K_{\text{w}}$ is,

$K_{\text{w}}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$

The value of $K_{\text{w}}$ is $1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2$.

The given ${\text{H}}_{\text{3}}{\text{O}}^{+}$ molar concentration is $3.3\times {10}^{-2}\text{mol}/\text{L}$. Substitute this value in the formula for ionic product.

$\begin{array}{rll}1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2& =& 3.3\times {10}^{-2}\text{mol}/\text{L}\left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]& =& \frac{1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2}{3.3\times {10}^{-2}\text{mol}/\text{L}}\\ \left[{\text{OH}}^{-}\right]& =& 3.03\times {10}^{-13}\text{mol}/\text{L}\end{array}$

Thus, the molar concentration of ${\text{OH}}^{-}$ in water solutions is $3.03\times {10}^{-13}\text{mol}/\text{L}$.

Conclusion

The molar concentration of ${\text{OH}}^{-}$ in water solutions is $3.03\times {10}^{-13}\text{mol}/\text{L}$.