Chapter 14, Problem 49E

Calculate the [OH^–] of each of the following solutions at 25°C. Identify each solution as neutral, acidic, or basic.

  a. [H⁺] = 1.0 × 10^–7 M

  b. [H⁺] = 8.3 × 10^–16 M

  c. [H⁺] = 12 M

  d. [H⁺] = 5.4 × 10^–5 M

  46. Calculate the [H⁺] of each of the following solutions at 25°C. Identify each solution as neutral, acidic, or basic.

  a. [OH^–] = 1.5 M

  b. [OH^–] = 3.6 × 10^–15 M

  c. [OH^–] = 1.0 × 10^–7 M

  d. [OH^–] = 7.3 × 10^–4 M

  49. Calculate the pH and pOH of the solutions in Exercises 45 and 46.

Interpretation Introduction

Interpretation:

The $\text{pH}$ and the $\text{pOH}$ values for thesolution in Exercises $45$ and $46$ are to be calculated.

Concept introduction:

The $\text{pH}$ of a solution is define as a figure that expresses the acidity of the alkalinity of a given solution. A logarithmic scale is used on which, the value $7$ corresponds to a neutral species, a value less than $7$ corresponds to an acid and a value greater than $7$ corresponds to a base.

The $\text{pOH}$ of a solution is the measure of the $\left[{\text{OH}}^{-}\right]$.

The $\text{pH}$ of a solution is calculated by the formula, $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

The $\text{pOH}$ of a solution is calculated by the formula, $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

At $25°\text{C}$, $\text{pH}+\text{pOH}=14$

Answer to Problem 49E

Answer

(45a)

The $\text{pH}$ value is $\underset{\_}{7}$ and the $\text{pOH}$ value is $\underset{\_}{7}$.

(45b)

The $\text{pH}$ value is $\underset{\_}{15}$ and the $\text{pOH}$ value is $\underset{\_}{-1}$.

(45c)

The $\text{pH}$ value is $\underset{\_}{1.1}$ and the $\text{pOH}$ value is $\underset{\_}{12.9}$.

(45d)

The $\text{pH}$ value is $\underset{\_}{4.3}$ and the $\text{pOH}$ value is $\underset{\_}{9.7}$.

(46a)

The $\text{pOH}$ value is $\underset{\_}{0.18}$ and the $\text{pOH}$ value is $\underset{\_}{13.82}$.

(46b)

The $\text{pOH}$ value is $\underset{\_}{14.4}$ and the $\text{pOH}$ value is $\underset{\_}{-0.4}$.

(46c)

The $\text{pOH}$ value is $\underset{\_}{7}$ and the $\text{pOH}$ value is $\underset{\_}{7}$.

(46d)

The $\text{pOH}$ value is $\underset{\_}{3.14}$ and the $\text{pOH}$ value is $\underset{\_}{10.86}$.

To determine: The $\text{pH}$ and the $\text{pOH}$ value for the given species.

Explanation of Solution

Explanation

(45a)

Calculation of $\text{pH}$ of the solution

The $\text{pH}$ value is $\underset{\_}{7}$.

Given

The temperature is $\text{25}\text{°C}$.

The $\left[{\text{H}}^{\text{+}}\right]=1.0\times {10}^{-7}\text{M}$.

The $\text{pH}$ of a solution is calculated by the formula,

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Substitute the given value of $\left[{\text{H}}^{+}\right]$ in the above expression.

$\begin{array}{c}\text{pH}=-\log \left[1.0\times {10}^{-7}\right]\\ =\underset{\_}{7}\end{array}$

Calculation of $\text{pOH}$ of the solution

The $\text{pOH}$ value is $\underset{\_}{7}$.

At $25°\text{C}$,

$\text{pH}+\text{pOH}=14$

Substitute the calculated value of $\text{pH}$ in the above expression.

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{7}+\text{pOH}=14\\ \text{pOH}=\underset{\_}{7}\end{array}$

(45b)

Calculation of $\text{pH}$ of the solution

The $\text{pH}$ value is $\underset{\_}{15}$.

Given

The temperature is $\text{25}\text{°C}$.

The $\left[{\text{H}}^{\text{+}}\right]=8.3\times {10}^{-16}\text{M}$.

The $\text{pH}$ of a solution is calculated by the formula,

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Substitute the given value of $\left[{\text{H}}^{+}\right]$ in the above expression.

$\begin{array}{c}\text{pH}=-\log \left[8.3\times {10}^{-16}\right]\\ =\underset{\_}{15}\end{array}$

Calculation of $\text{pOH}$ of the solution

The $\text{pOH}$ value is $\underset{\_}{-1}$.

At $25°\text{C}$,

$\text{pH}+\text{pOH}=14$

Substitute the calculated value of $\text{pH}$ in the above expression.

$\begin{array}{c}\text{pH}+\text{pOH}=15\\ \text{15}+\text{pOH}=14\\ \text{pOH}=\underset{\_}{-1}\end{array}$

(45c)

The $\text{pH}$ value is $\underset{\_}{1.1}$.

Given

The temperature is $\text{25}\text{°C}$.

The $\left[{\text{H}}^{\text{+}}\right]=12\text{M}$.

The $\text{pH}$ of a solution is calculated by the formula,

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Substitute the given value of $\left[{\text{H}}^{+}\right]$ in the above expression.

$\begin{array}{c}\text{pH}=-\log \left[12\right]\\ \approx \underset{\_}{1.1}\end{array}$

The $\text{pOH}$ value is $\underset{\_}{12.9}$.

At $25°\text{C}$,

$\text{pH}+\text{pOH}=14$

Substitute the calculated value of $\text{pH}$ in the above expression.

$\begin{array}{c}\text{pH}+\text{pOH}=\text{1}\text{.1}\\ \text{1}\text{.1}+\text{pOH}=14\\ \text{pOH}=\underset{\_}{12.9}\end{array}$

(45d)

The $\text{pH}$ value is $\underset{\_}{4.3}$.

Given

The temperature is $\text{25}\text{°C}$.

The $\left[{\text{H}}^{\text{+}}\right]=5.4\times {10}^{-5}\text{M}$.

The $\text{pH}$ of a solution is calculated by the formula,

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Substitute the given value of $\left[{\text{H}}^{+}\right]$ in the above expression.

$\begin{array}{c}\text{pH}=-\log \left[5.4\times {10}^{-5}\right]\\ \approx \underset{\_}{4.3}\end{array}$

The $\text{pOH}$ value is $\underset{\_}{9.7}$.

At $25°\text{C}$,

$\text{pH}+\text{pOH}=14$

Substitute the calculated value of $\text{pH}$ in the above expression.

$\begin{array}{c}\text{pH}+\text{pOH}=\text{4}\text{.3}\\ \text{4}\text{.3}+\text{pOH}=14\\ \text{pOH}=\underset{\_}{9.7}\end{array}$

(46a)

The $\text{pOH}$ value is $\underset{\_}{0.18}$.

Given

The temperature is $\text{25}\text{°C}$.

The $\left[{\text{OH}}^{-}\right]=1.5\text{M}$.

The $\text{pOH}$ of a solution is calculated by the formula,

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Substitute the given value of $\left[{\text{OH}}^{-}\right]$ in the above expression.

$\begin{array}{c}\text{pOH}=-\log \left[1.5\right]\\ =\underset{\_}{0.18}\end{array}$

The $\text{pOH}$ value is $\underset{\_}{13.82}$.

At $25°\text{C}$,

$\text{pH}+\text{pOH}=14$

Substitute the calculated value of $\text{pOH}$ in the above expression.

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ 0.18+\text{pH}=14\\ \text{pOH}=\underset{\_}{13.82}\end{array}$

(46b)

The $\text{pOH}$ value is $\underset{\_}{14.4}$.

Given

The temperature is $\text{25}\text{°C}$.

The $\left[{\text{OH}}^{-}\right]=3.6\times {10}^{-15}\text{M}$.

The $\text{pOH}$ of a solution is calculated by the formula,

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Substitute the given value of $\left[{\text{OH}}^{-}\right]$ in the above expression.

$\begin{array}{c}\text{pOH}=-\log \left[3.6\times {10}^{-15}\right]\\ =\underset{\_}{14.4}\end{array}$

The $\text{pOH}$ value is $\underset{\_}{-0.4}$.

At $25°\text{C}$,

$\text{pH}+\text{pOH}=14$

Substitute the calculated value of $\text{pOH}$ in the above expression.

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ 14.4+\text{pH}=14\\ \text{pOH}=\underset{\_}{-0.4}\end{array}$

(46c)

The $\text{pOH}$ value is $\underset{\_}{7}$.

Given

The temperature is $\text{25}\text{°C}$.

The $\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-7}\text{M}$.

The $\text{pOH}$ of a solution is calculated by the formula,

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Substitute the given value of $\left[{\text{OH}}^{-}\right]$ in the above expression.

$\begin{array}{c}\text{pOH}=-\log \left[1.0\times {10}^{-7}\right]\\ =\underset{\_}{7}\end{array}$

The $\text{pOH}$ value is $\underset{\_}{7}$.

At $25°\text{C}$,

$\text{pH}+\text{pOH}=14$

Substitute the calculated value of $\text{pOH}$ in the above expression.

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ 7+\text{pH}=14\\ \text{pOH}=\underset{\_}{7}\end{array}$

(46d)

The $\text{pOH}$ value is $\underset{\_}{3.14}$.

Given

The temperature is $\text{25}\text{°C}$.

The $\left[{\text{OH}}^{-}\right]=7.3\times {10}^{-4}\text{M}$.

The $\text{pOH}$ of a solution is calculated by the formula,

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Substitute the given value of $\left[{\text{OH}}^{-}\right]$ in the above expression.

$\begin{array}{c}\text{pOH}=-\log \left[7.3\times {10}^{-4}\right]\\ =\underset{\_}{3.14}\end{array}$

The $\text{pOH}$ value is $\underset{\_}{10.86}$.

At $25°\text{C}$,

$\text{pH}+\text{pOH}=14$

Substitute the calculated value of $\text{pOH}$ in the above expression.

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ 3.14+\text{pH}=14\\ \text{pOH}=\underset{\_}{10.86}\end{array}$

Conclusion

At a temperature of $25°\text{C}$ $\text{pH and pOH}$ of the given solutions were calculated.