Chapter 17, Problem 30P

Interpretation Introduction

(a)

Interpretation:

The ratio of A- to [HA] with the help of pH and pK_a values should be calculated.

Concept Introduction:

The pH of a solution indicates the number of H₃ O+ ions in the solution. The mathematical relation between pH and H₃ O+ is as given below;.

$\begin{array}{l}\text{pH }= -\text{ log }\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\hfill \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right] ={10}^{-\text{pH}}\hfill \end{array}$

Acid dissociation constant K_a is related to pK_a as;.

$\begin{array}{l}p{K}_{\text{a}} = -\text{ log }{K}_{\text{a}}\hfill \\ K_{\text{a}}={10}^{-p{K}_{\text{a}}}\hfill \end{array}$.

Answer to Problem 30P

${10}^{-3}$.

Explanation of Solution

Given Information:

pKa = 5.0.

pH = 2.0.

Calculation:

From pH of solution, concentration of hydroxide ion can be calculated as follows:

$\begin{array}{l}\left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-p\text{H}}\\ ={10}^{-2 } \\ =0.01 \end{array}$

Also,

$\begin{array}{l} K_{\text{a}}={10}^{-pK\text{a}}\\ ={10}^{-5} \end{array}$

Since,

$\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}=\frac{K_{\text{a}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}$

Thus, by putting the values, the ratio $\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$ can be calculated as follows:

$\begin{array}{c}\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}=\frac{{10}^{-5}}{{10}^{-2}}\\ ={10}^{-3}\end{array}$.

Interpretation Introduction

(b)

Interpretation:

The ratio of A- to [HA] with the help of pH and pK_a values should be calculated.

Concept Introduction:

The pH of a solution indicates the number of H₃ O+ ions in the solution. The mathematical relation between pH and H₃ O+ is as given below;.

$\begin{array}{l}\text{pH }= -\text{ log }\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\hfill \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right] ={10}^{-\text{pH}}\hfill \end{array}$

Acid dissociation constant K_a is related to pK_a as;.

$\begin{array}{l}p{K}_{\text{a}} = -\text{ log }{K}_{\text{a}}\hfill \\ K_{\text{a}}={10}^{-p{K}_{\text{a}}}\hfill \end{array}$.

Answer to Problem 30P

Given:

pK_a = 5.0.

pH = 5.0.

Hence [H₃ O⁺ ] = 10-pH = 10-5.

And Ka = 10-pKa = 10-5.

Since

Plug the values of K_a and H₃ O+ ions to calculate the ratio of [A^- ] to [HA].

[A^- ] / [HA] = 10-5 / 10-5 = 1.

Explanation of Solution

Given Information:

pKa = 5.0.

pH = 5.0.

Calculation:

From pH of solution, concentration of hydroxide ion can be calculated as follows:

$\begin{array}{l}\left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-p\text{H}}\\ ={10}^{-5 } \\ =1.0\times {10}^{-5 } \end{array}$

Also,

$\begin{array}{l} K_{\text{a}}={10}^{-pK\text{a}}\\ ={10}^{-5} \end{array}$

Since,

$\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}=\frac{K_{\text{a}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}$

Thus, by putting the values, the ratio $\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$ can be calculated as follows:

$\begin{array}{c}\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}=\frac{{10}^{-5}}{{10}^{-5}}\\ =1\end{array}$.

Interpretation Introduction

(c)

Interpretation:

The ratio of A- to [HA] with the help of pH and pK_a values should be calculated.

Concept Introduction:

The pH of a solution indicates the number of H₃ O+ ions in the solution. The mathematical relation between pH and H₃ O+ is as given below;.

$\begin{array}{l}\text{pH }= -\text{ log }\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\hfill \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right] ={10}^{-\text{pH}}\hfill \end{array}$

Acid dissociation constant K_a is related to pK_a as;.

$\begin{array}{l}p{K}_{\text{a}} = -\text{ log }{K}_{\text{a}}\hfill \\ K_{\text{a}}={10}^{-p{K}_{\text{a}}}\hfill \end{array}$.

Answer to Problem 30P

${10}^2$.

Explanation of Solution

Given Information:

pKa = 5.0.

pH = 7.0.

Calculation:

From pH of solution, concentration of hydroxide ion can be calculated as follows:

$\begin{array}{l}\left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-p\text{H}}\\ ={10}^{-7 } \\ =1.0\times {10}^{-7 } \end{array}$

Also,

$\begin{array}{l} K_{\text{a}}={10}^{-pK\text{a}}\\ ={10}^{-5} \end{array}$

Since,

$\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}=\frac{K_{\text{a}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}$

Thus, by putting the values, the ratio $\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$ can be calculated as follows:

$\begin{array}{c}\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}=\frac{{10}^{-5}}{1.0\times {10}^{-7 } }\\ ={10}^2\end{array}$.

Interpretation Introduction

(d)

Interpretation:

The ratio of A- to [HA] with the help of pH and pK_a values should be calculated.

Concept Introduction:

The pH of a solution indicates the number of H₃ O+ ions in the solution. The mathematical relation between pH and H₃ O+ is as given below;.

$\begin{array}{l}\text{pH }= -\text{ log }\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\hfill \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right] ={10}^{-\text{pH}}\hfill \end{array}$

Acid dissociation constant K_a is related to pK_a as;.

$\begin{array}{l}p{K}_{\text{a}} = -\text{ log }{K}_{\text{a}}\hfill \\ K_{\text{a}}={10}^{-p{K}_{\text{a}}}\hfill \end{array}$.

Answer to Problem 30P

${10}^4$.

Explanation of Solution

Given Information:

pKa = 5.0.

pH = 9.0.

Calculation:

From pH of solution, concentration of hydroxide ion can be calculated as follows:

$\begin{array}{l}\left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-p\text{H}}\\ ={10}^{-9 } \\ =1.0\times {10}^{-9 }\end{array}$

Also,

$\begin{array}{l} K_{\text{a}}={10}^{-pK\text{a}}\\ ={10}^{-5} \end{array}$

Since,

$\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}=\frac{K_{\text{a}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}$

Thus, by putting the values, the ratio $\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$ can be calculated as follows:

$\begin{array}{c}\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}=\frac{{10}^{-5}}{1.0\times {10}^{-9 } }\\ ={10}^4\end{array}$.

Interpretation Introduction

(e)

Interpretation:

The ratio of A- to [HA] with the help of pH and pK_a values should be calculated.

Concept Introduction:

The pH of a solution indicates the number of H₃ O+ ions in the solution. The mathematical relation between pH and H₃ O+ is as given below;.

$\begin{array}{l}\text{pH }= -\text{ log }\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\hfill \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right] ={10}^{-\text{pH}}\hfill \end{array}$

Acid dissociation constant K_a is related to pK_a as;.

$\begin{array}{l}p{K}_{\text{a}} = -\text{ log }{K}_{\text{a}}\hfill \\ K_{\text{a}}={10}^{-p{K}_{\text{a}}}\hfill \end{array}$.

Answer to Problem 30P

${10}^6$.

Explanation of Solution

Given Information:

pKa = 5.0.

pH = 11.0.

Calculation:

From pH of solution, concentration of hydroxide ion can be calculated as follows:

$\begin{array}{l}\left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-p\text{H}}\\ ={10}^{-11 } \\ =1.0\times {10}^{-11 }\end{array}$

Also,

$\begin{array}{l} K_{\text{a}}={10}^{-pK\text{a}}\\ ={10}^{-5} \end{array}$

Since,

$\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}=\frac{K_{\text{a}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}$

Thus, by putting the values, the ratio $\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$ can be calculated as follows:

$\begin{array}{c}\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}=\frac{{10}^{-5}}{1.0\times {10}^{-11 } }\\ ={10}^6\end{array}$.