Chapter 16, Problem 16.130QP

Malic acid is a weak diprotic organic acid with K_a1 = 4.0 × 10⁻⁴ and K_a2 = 9.0 × 10⁻⁵.

1. a Letting the symbol H₂A represent malic acid, write the chemical equations that represent K_a1 and K_a2. Write the chemical equation that represents K_a1 × K_a2.
2. b Qualitatively describe the relative concentrations of H₂A, HA⁻, A²⁻, and H₃O⁺ in a solution that is about one molar in malic acid.
3. c Calculate the pH of a 0.0175 M malic acid solution and the equilibrium concentration of [H₂A].
4. d What is the A²⁻ concentrationin in solutions b and c?

Interpretation Introduction

Interpretation:

Malic acid is weak diprotic acid with ${\text{K}}_{a1}=4.0\times {10}^{-4}$ and ${\text{K}}_{a2}=9.0\times {10}^{-5}$

Considering the symbol ${\text{H}}_{\text{2}}\text{A}$ representing the malic acid, the chemical equations that represents ${\text{K}}_{a1}$ and ${\text{K}}_{a2}$ has to be written.  Also, the chemical equation that represents ${\text{K}}_{\text{a1}}\times {\text{K}}_{\text{a2}}$ has to be written

Concept Introduction:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The ionization of a weak acid $\text{HA}$ can be given as follows,

${\text{HA}}_{(aq)}\rightleftarrows {\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$

The equilibrium expression for the above reaction is given below.

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

Where,

$K_a$ is acid ionization constant,

$[{\text{H}}^{\text{+}}\text{]}$ is concentration of hydrogen ion

$[{\text{A}}^{\text{-}}\text{]}$ is concentration of acid anion

$[\text{HA]}$ is concentration of the acid

Diprotic and polyprotic acids:

Acids having two or more hydrogen atoms are termed as diprotic or polyprotic acids.  These acids lose one proton at a time by undergoing successive ionizations.

For diprotic acids, the successive ionization constants are designated as $K_{a1}and{K}_{a2}$

For triprotic acids, the successive ionization constants are designated as $K_{a1},K_{a2}and{K}_{a3}$

Answer to Problem 16.130QP

(a)

The chemical equation representing ${\text{K}}_{a1}$ is: ${\text{H}}_{\text{2}}{\text{A + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ + HA}}^{\text{-}}$

The chemical equation representing ${\text{K}}_{a2}$ is: ${\text{HA}}^{\text{-}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ + A}}^{\text{2-}}$

The chemical equation representing ${\text{K}}_{\text{a1}}\times {\text{K}}_{\text{a2}}$ is: ${\text{H}}_{\text{2}}{\text{A + 2H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ 2H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ + A}}^{\text{2-}}$

Explanation of Solution

To Write: Considering the symbol ${\text{H}}_{\text{2}}\text{A}$ representing the malic acid, the chemical equations that represents ${\text{K}}_{a1}$ and ${\text{K}}_{a2}$ and also the chemical equation that represents ${\text{K}}_{\text{a1}}\times {\text{K}}_{\text{a2}}$

Given data:

Malic acid is a weak diprotic organic acid with ${\text{K}}_{a1}=4.0\times {10}^{-4}$ and ${\text{K}}_{a2}=9.0\times {10}^{-6}$

The chemical equation representing ${\text{K}}_{a1}$ is:

${\text{H}}_{\text{2}}{\text{A + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ + HA}}^{\text{-}}$

The chemical equation representing ${\text{K}}_{a2}$ is:

${\text{HA}}^{\text{-}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ + A}}^{\text{2-}}$

The chemical equation representing ${\text{K}}_{\text{a1}}\times {\text{K}}_{\text{a2}}$ is:

$\begin{array}{l}\text{K}\text{=}{\text{K}}_{\text{a1}}\times {\text{K}}_{\text{a2}}\\ {\text{H}}_{\text{2}}{\text{A + 2H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ 2H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ + A}}^{\text{2-}}\end{array}$

Conclusion

The chemical equation representing ${\text{K}}_{a1}$ was written as: ${\text{H}}_{\text{2}}{\text{A + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ + HA}}^{\text{-}}$

The chemical equation representing ${\text{K}}_{a2}$ was written as: ${\text{HA}}^{\text{-}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ + A}}^{\text{2-}}$

The chemical equation representing ${\text{K}}_{\text{a1}}\times {\text{K}}_{\text{a2}}$ was written as: ${\text{H}}_{\text{2}}{\text{A + 2H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ 2H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ + A}}^{\text{2-}}$

Interpretation Introduction

Interpretation:

Malic acid is weak diprotic acid with ${\text{K}}_{a1}=4.0\times {10}^{-4}$ and ${\text{K}}_{a2}=9.0\times {10}^{-5}$

Qualitatively the relative concentrations of ${\text{H}}_2\text{A}$, ${\text{HA}}^{-}$, ${\text{A}}^{\text{2-}}$ and ${\text{H}}_{\text{3}}{\text{O}}^{+}$ in a one molar solution of malic acid has to be given

Concept Introduction:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The ionization of a weak acid $\text{HA}$ can be given as follows,

${\text{HA}}_{(aq)}\rightleftarrows {\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$

The equilibrium expression for the above reaction is given below.

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

Where,

$K_a$ is acid ionization constant,

$[{\text{H}}^{\text{+}}\text{]}$ is concentration of hydrogen ion

$[{\text{A}}^{\text{-}}\text{]}$ is concentration of acid anion

$[\text{HA]}$ is concentration of the acid

Diprotic and polyprotic acids:

Acids having two or more hydrogen atoms are termed as diprotic or polyprotic acids.  These acids lose one proton at a time by undergoing successive ionizations.

For diprotic acids, the successive ionization constants are designated as $K_{a1}and{K}_{a2}$

For triprotic acids, the successive ionization constants are designated as $K_{a1},K_{a2}and{K}_{a3}$

Answer to Problem 16.130QP

The relative concentrations of ${\text{H}}_2\text{A}$, ${\text{HA}}^{-}$, ${\text{A}}^{\text{2-}}$ and ${\text{H}}_{\text{3}}{\text{O}}^{+}$ in a 0.5 M tartaric  acid solution is: ${\text{H}}_{\text{2}}{\text{A >> H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ = HA}}^{\text{-}}{\text{ >> A}}^{\text{2-}}$

Explanation of Solution

To Give: Qualitatively the relative concentrations of ${\text{H}}_2\text{A}$, ${\text{HA}}^{-}$, ${\text{A}}^{\text{2-}}$ and ${\text{H}}_{\text{3}}{\text{O}}^{+}$ in a one molar solution of malic acid

The relative concentrations of ${\text{H}}_2\text{A}$, ${\text{HA}}^{-}$, ${\text{A}}^{\text{2-}}$ and ${\text{H}}_{\text{3}}{\text{O}}^{+}$ in a one molar solution of malic acid is:

${\text{H}}_{\text{2}}{\text{A >> H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ = HA}}^{\text{-}}{\text{ >> A}}^{\text{2-}}$

Conclusion

The relative concentrations of ${\text{H}}_2\text{A}$, ${\text{HA}}^{-}$, ${\text{A}}^{\text{2-}}$ and ${\text{H}}_{\text{3}}{\text{O}}^{+}$ in a one molar solution of malic acid was given as: ${\text{H}}_{\text{2}}{\text{A >> H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ = HA}}^{\text{-}}{\text{ >> A}}^{\text{2-}}$

Interpretation Introduction

Interpretation:

Malic acid is weak diprotic acid with ${\text{K}}_{a1}=4.0\times {10}^{-4}$ and ${\text{K}}_{a2}=9.0\times {10}^{-5}$

The pH of 0.0175 M malic acid solution and the equilibrium concentration of ${\text{[H}}_{\text{2}}\text{A]}$ has to be calculated

Concept Introduction:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The ionization of a weak acid $\text{HA}$ can be given as follows,

${\text{HA}}_{(aq)}\rightleftarrows {\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$

The equilibrium expression for the above reaction is given below.

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

Where,

$K_a$ is acid ionization constant,

$[{\text{H}}^{\text{+}}\text{]}$ is concentration of hydrogen ion

$[{\text{A}}^{\text{-}}\text{]}$ is concentration of acid anion

$[\text{HA]}$ is concentration of the acid

Diprotic and polyprotic acids:

Acids having two or more hydrogen atoms are termed as diprotic or polyprotic acids.  These acids lose one proton at a time by undergoing successive ionizations.

For diprotic acids, the successive ionization constants are designated as $K_{a1}and{K}_{a2}$

For triprotic acids, the successive ionization constants are designated as $K_{a1},K_{a2}and{K}_{a3}$

Answer to Problem 16.130QP

The pH of 0.0175 M malic acid solution is 2.16

The equilibrium concentration of ${\text{[H}}_{\text{2}}\text{A]}$ is 0.0150 M

Explanation of Solution

To Calculate: The pH of 0.0175 M malic acid solution and the equilibrium concentration of ${\text{[H}}_{\text{2}}\text{A]}$

The reaction is:

$\begin{array}{l}{\text{H}}_{\text{2}}\text{A }\text{ + }{\text{H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ + HA}}^{\text{-}}\\ \text{0}\text{.0175-}xxx\end{array}$

The ${\text{K}}_{\text{a1}}$ value is $K_{a1}=4.0\times {10}^{-4}$

Substitute into the equilibrium constant expression

$\begin{array}{l}{\text{K}}_{a1}=\frac{{\text{[HA}}^{-}][{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[H}}_{\text{2}}\text{A]}}\\ 4.0\times {10}^{-4}=\frac{x^2}{(0.0175-x)}\\ \text{Solving }x\text{ using quadratic equation, we get}\\ x^2+(4.0\times {10}^{-4})x+(-7.00\times {10}^{-6})=0\\ x=\frac{-(4.0\times {10}^{-4})\pm \sqrt{{(4.0\times {10}^{-4})}^2-4(1)(-7.00\times {10}^{-6})}}{2}\\ x=2.5\times {10}^{-3}\text{ M}\end{array}$

The pH is calculated as follows,

$\begin{array}{l}\text{pH}{\text{=-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{=-log(}2.5\times {10}^{-3})\\ =2.61\end{array}$

The concentration of ${\text{[H}}_{\text{2}}\text{A]}$ is:

$\begin{array}{l}{\text{[H}}_{\text{2}}\text{A]}\text{=0}\text{.0175 - x}\\ \text{=0}\text{.0175 - }2.5\times {10}^{-3}\\ =0.0150\text{ M}\end{array}$

Conclusion

The pH of 0.0175 M malic acid solution is 2.61

The equilibrium concentration of ${\text{[H}}_{\text{2}}\text{A]}$ is 0.0150 M

Interpretation Introduction

Interpretation:

Malic acid is weak diprotic acid with ${\text{K}}_{a1}=4.0\times {10}^{-4}$ and ${\text{K}}_{a2}=9.0\times {10}^{-5}$

The ${\text{A}}^{\text{2-}}$ concentration in solutions (b) and (c) has to be calculated

Concept Introduction:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The ionization of a weak acid $\text{HA}$ can be given as follows,

${\text{HA}}_{(aq)}\rightleftarrows {\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$

The equilibrium expression for the above reaction is given below.

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

Where,

$K_a$ is acid ionization constant,

$[{\text{H}}^{\text{+}}\text{]}$ is concentration of hydrogen ion

$[{\text{A}}^{\text{-}}\text{]}$ is concentration of acid anion

$[\text{HA]}$ is concentration of the acid

Diprotic and polyprotic acids:

Acids having two or more hydrogen atoms are termed as diprotic or polyprotic acids.  These acids lose one proton at a time by undergoing successive ionizations.

For diprotic acids, the successive ionization constants are designated as $K_{a1}and{K}_{a2}$

For triprotic acids, the successive ionization constants are designated as $K_{a1},K_{a2}and{K}_{a3}$

Answer to Problem 16.130QP

The ${\text{A}}^{\text{2-}}$ concentration in solutions (b) and (c) is $9.0\times {10}^{-6}\text{ M}$

Explanation of Solution

To Calculate: The ${\text{A}}^{\text{2-}}$ concentration in solutions (b) and (c)

Concentration of ${\text{A}}^{\text{2-}}$in solution (c):

Consider the second ionization of the acid for the calculation of ${\text{A}}^{\text{2-}}$ concentration.

$\begin{array}{l}{\text{ HA}}^{-}+{\text{ H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ H}}_{\text{3}}{\text{O}}^{+}\text{+ }{\text{A}}^{2-}\\ 2.453\times {10}^{-3}+y2.453\times {10}^{-3}+yy\end{array}$

The ${\text{K}}_{\text{a2}}$ value is $K_{a2}=9.0\times {10}^{-6}$

Substitute into the equilibrium constant expression

Assume y is small compared to $2.453\times {10}^{-3}$ and neglect it.

$\begin{array}{l}{\text{[A}}^{\text{2-}}\text{]}=\frac{{\text{[HA}}^{-}][{\text{K}}_{a2}\text{]}}{{\text{[H}}_3{\text{O}}^{+}\text{]}}\\ =\frac{(2.453\times {10}^{-3})(9.0\times {10}^{-6})}{(2.453\times {10}^{-3})}\\ x=9.0\times {10}^{-6}\text{ M}\end{array}$

Similarly, the concentration of ${\text{A}}^{\text{2-}}$ in solution (b) is also $9.0\times {10}^{-6}\text{ M}$

Conclusion

The ${\text{A}}^{\text{2-}}$ concentration in solutions (b) and (c) was calculated as $9.0\times {10}^{-6}\text{ M}$