Chapter 14, Problem 9QAP

Calculate [H⁺] and pH in a solution in which lactic acid, HC₃H₅O₃, is 0.250 M and the lactate ion, C₃H₅O₃^-, is

(a) 0.250 M (b) 0.125 M

(c) 0.0800 M (d) 0.0500 M

Interpretation Introduction

Interpretation:

The $\left[{\text{H}}^{+}\right]$ and pH of the solution with molarity of lactic acid 0.50 M and lactate ion 0.250 M needs to be determined.

Concept Introduction :

For an acid dissociation reaction, the $\left[{\text{H}}^{+}\right]$ can be calculated from acid dissociation constant as follows:

  $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$

Here,

  $K_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

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

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

Answer to Problem 9QAP

  $1.4\times {10}^{-4}$ M and 3.85

Explanation of Solution

The equilibrium reaction is as follows:

  $H{C}_3{H}_5{O}_3\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+C_3{H}_5{O}_3{}^{-}\left(aq\right)$

The expression of K_a for this reaction is as follows:

  ${\text{K}}_{\text{a}}\text{= }\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{\text{-}}\right]}{\left[{\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}\right]}$

The above expression can be written as follows:

  $\left[H^{+}\right]=\frac{K_a\left[{}^{}{C}_3{H}_5{O}_3{}^{-}\right]}{\left[H{C}_3{H}_5{O}_3\right]}$

The K_a value of lactic acid HC₃H₅O₃ is 1.4 x 10-4, the concentrated of lactic acid (HC₃H₅O₃) is 0.250 M, and for lactate ion C₃H₅O₃- is 0.250 M.

Substitute these values in the above expression as follows:

  $\left[H^{+}\right]=\frac{K_a\left[{}^{}{C}_3{H}_5{O}_3{}^{-}\right]}{\left[H{C}_3{H}_5{O}_3\right]}$

  = $\frac{\left(1.4x{10}^{-4}\right)\left(0.250M\right)}{\left(0.250M\right)}$

  = $1.4\times {10}^{-4}$

Now, the pH of the solution can be calculated by using the [H⁺] as follows:

  $\begin{array}{l}\text{pH }=-lo{g}_{10}\left[{\text{H}}^{+}\right]\hfill \\ =-{\text{log}}_{10}\left[1.4\times {10}^{-4}\right]\hfill \\ =3.85\hfill \end{array}$

So, the pH of the solution is 3.85 and $\left[H^{+}\right]$ is $1.4\times {10}^{-4}$ M.

Interpretation Introduction

Interpretation:

The $\left[{\text{H}}^{+}\right]$ and pH of the solution with molarity of lactic acid 0.50 M and lactate ion 0.125 M needs to be determined.

Concept Introduction :

For an acid dissociation reaction, the $\left[{\text{H}}^{+}\right]$ can be calculated from acid dissociation constant as follows:

  $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$

Here,

  $K_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

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

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

Answer to Problem 9QAP

  $2.8\times {10}^{-4}$ M and 3.55.

Explanation of Solution

The equilibrium reaction is as follows:

  $H{C}_3{H}_5{O}_3\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+C_3{H}_5{O}_3{}^{-}\left(aq\right)$

The expression of K_a for this reaction is as follows:

  ${\text{K}}_{\text{a}}\text{= }\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{\text{-}}\right]}{\left[{\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}\right]}$

The above expression can be written as follows:

  $\left[H^{+}\right]=\frac{K_a\left[{}^{}{C}_3{H}_5{O}_3{}^{-}\right]}{\left[H{C}_3{H}_5{O}_3\right]}$

The K_a value of lactic acid HC₃H₅O₃ is 1.4 x 10-4, the concentrated of lactic acid (HC₃H₅O₃) is 0.250 M, and for lactate ion C₃H₅O₃- is 0.125 M.

Substitute these values in the above expression as follows:

  $\left[H^{+}\right]=\frac{K_a\left[{}^{}{C}_3{H}_5{O}_3{}^{-}\right]}{\left[H{C}_3{H}_5{O}_3\right]}$

  = $\frac{\left(1.4x{10}^{-4}\right)\left(0.250M\right)}{\left(0.125M\right)}$

  = $2.8\times {10}^{-4}$

Now, the pH of the solution can be calculated by using the [H⁺] as follows:

pH = -log₁₀ [H⁺]

= -log₁₀ [2.8 x 10-4]

= 3.55

So, the pH of the solution is 3.55 and [H⁺] is $2.8\times {10}^{-4}$ M.

Interpretation Introduction

Interpretation:

The $\left[{\text{H}}^{+}\right]$ and pH of the solution with molarity of lactic acid 0.50 M and lactate ion 0.08 M needs to be determined.

Concept Introduction :

For an acid dissociation reaction, the $\left[{\text{H}}^{+}\right]$ can be calculated from acid dissociation constant as follows:

  $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$

Here,

  $K_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

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

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

Answer to Problem 9QAP

  $4.4\times {10}^{-4}$ M and 3.36

Explanation of Solution

The equilibrium reaction is as follows:

  $H{C}_3{H}_5{O}_3\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+C_3{H}_5{O}_3{}^{-}\left(aq\right)$

The expression of K_a for this reaction is as follows:

  ${\text{K}}_{\text{a}}\text{= }\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{\text{-}}\right]}{\left[{\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}\right]}$

The above expression can be written as follows:

  $\left[H^{+}\right]=\frac{K_a\left[{}^{}{C}_3{H}_5{O}_3{}^{-}\right]}{\left[H{C}_3{H}_5{O}_3\right]}$

The K_a value of lactic acid HC₃H₅O₃ is 1.4 x 10-4, the concentrated of lactic acid (HC₃H₅O₃) is 0.250 M, and for lactate ion C₃H₅O₃- is 0.0800 M.

Substitute these values in the above expression as follows:

  $\left[H^{+}\right]=\frac{K_a\left[{}^{}{C}_3{H}_5{O}_3{}^{-}\right]}{\left[H{C}_3{H}_5{O}_3\right]}$

  = $\frac{\left(1.4x{10}^{-4}\right)\left(0.125M\right)}{\left(0.0800M\right)}$

  = $4.4\times {10}^{-4}$

Now, the pH of the solution can be calculated by using the [H⁺] as follows:

pH = -log₁₀ [H⁺]

= -log₁₀$4.4\times {10}^{-4}$

= 3.36

So, the pH of the solution is 3.36 and [H⁺] is $4.4\times {10}^{-4}$ M.

Interpretation Introduction

Interpretation:

The $\left[{\text{H}}^{+}\right]$ and pH of the solution with molarity of lactic acid 0.50 M and lactate ion 0.0500 M needs to be determined.

Concept Introduction :

For an acid dissociation reaction, the $\left[{\text{H}}^{+}\right]$ can be calculated from acid dissociation constant as follows:

  $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$

Here,

  $K_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

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

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

Answer to Problem 9QAP

  $7\times {10}^{-4}$ M and 3.15.

Explanation of Solution

The equilibrium reaction is as follows:

  $H{C}_3{H}_5{O}_3\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+C_3{H}_5{O}_3{}^{-}\left(aq\right)$

The expression of K_a for this reaction is as follows:

  ${\text{K}}_{\text{a}}\text{= }\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{\text{-}}\right]}{\left[{\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}\right]}$

The above expression can be written as follows:

  $\left[H^{+}\right]=\frac{K_a\left[{}^{}{C}_3{H}_5{O}_3{}^{-}\right]}{\left[H{C}_3{H}_5{O}_3\right]}$

The K_a value of lactic acid HC₃H₅O₃ is 1.4 x 10-4, the concentrated of lactic acid (HC₃H₅O₃) is 0.250 M, and for lactate ion C₃H₅O₃- is 0.0500 M.

Substitute these values in the above expression as follows:

  $\left[H^{+}\right]=\frac{K_a\left[{}^{}{C}_3{H}_5{O}_3{}^{-}\right]}{\left[H{C}_3{H}_5{O}_3\right]}$

  = $\frac{\left(1.4x{10}^{-4}\right)\left(0.125M\right)}{\left(0.0500M\right)}$

  = $7\times {10}^{-4}$

Now, the pH of the solution can be calculated by using the [H⁺] as follows:

pH = -log₁₀ [H⁺]

= -log₁₀$7\times {10}^{-4}$

= 3.15

So, the pH of the solution is 3.15 and [OH^-] is $7\times {10}^{-4}$ M.