Chapter 17, Problem 1E

For a solution that e 0.275M ${\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}$ COOH (propionic acid, $K_8=1.3\times {10}^{-9}$ ) and 0.0892 M HI, calculate (a) [HO^-]; (b) [OH^-] (c) CH₂CH₂COO; (d) [1^-].

Interpretation Introduction

(a)

Interpretation:

A solution consists propionic acid, ${\mathrm{CH}}_{3}C{H}_{2}COOH$ of molarity 0.275 M and HI of molarity 0.0892 M. The concentration of $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$ is to be calculated. The value of equilibrium constant of acid is $K_a=1.3\times {10}^{-5}$.

Concept introduction:

The concentration of any solution is calculated by the molarity of the solution. It is defined as the number of moles of solute in 1 L of the solution.

$M=\frac{n}{V\left(\text{L}\right)}$

Here, n is number of moles of solute and V is volume of solution in L.

The dissociation constant of the weak acid is calculated by the following formula-

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

The expression can be written as follows:

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

Answer to Problem 1E

The concentration of $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$ = 0.0892 M.

Explanation of Solution

The complete dissociation of strong acid occurs in the solution, whereas weak acid is not able to dissociate completely.

From the given information,

Propionic acid, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}$ or ${\mathrm{C}}_3{H}_6{O}_2$ is a weak acid.

The molarity of propionic acid $\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}\right]$ is 0.275 M.

Hydroiodic acid, HI = strong acid

The molarity of [HI] or [H₃O⁺] ion = 0.0892 M

The chemical equation of this reaction is given below:

${\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}(aq)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_2^{-}(aq)+{\text{H}}_{\text{3}}{\text{O}}^{+}(aq)$

Now, the concentration will be calculated by using the following ICE table:

	${\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}$	${\text{H}}_{\text{2}}\text{O}$	${\mathrm{C}}_3{H}_5{O}_2^{-}$	${\text{H}}_{\text{3}}{\text{O}}^{+}$
Initial (I)	0.275	-	0	0.0892
Change (C)	- x	-	+ x	+ x
Equilibrium (E)	0.275- x	-	x	0.0892+ x

Given that-

K_a = 1.3×10-5

${\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_2^{-}$ = x

${\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}$ = 0.275-x

${\text{H}}_{\text{3}}{\text{O}}^{+}$ = 0.0892+x

The equation for the K_a is as follows:

$K_a=\frac{{\text{[C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[C}}_{\text{3}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}\text{]}}$ …………… (1)

Put the given values in equation (1).

$1.3\times {10}^{-5}=\frac{x\left(0.0892+x\right)}{0.275-x}$

On calculation −

$x=4.0\times {10}^{-5}M$

Now, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$

$= 0.0892+x=0.0892+4.0\times {10}^{-5}$

${\mathrm{H}}_3{O}^{+}\approx 0.089\text{2 M}$

Interpretation Introduction

(b)

Interpretation:

A solution consists propionic acid, ${\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{COOH}$ of molarity 0.275 M and HI of molarity 0.0892 M. The concentration of $\left[{\text{OH}}^{-}\right]$ is to be calculated. The value of equilibrium constant of acid is $K_a=1.3\times {10}^{-3}$.

Concept introduction:

The concentration of any solution is calculated by the molarity of the solution. It is defined as the number of moles of solute in 1 L of the solution.

$M=\frac{n}{V\left(\text{L}\right)}$

Here, n is number of moles of solute and V is volume of solution in L.

The dissociation constant of the weak acid is calculated by the following formula:

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

The expression can be written as follows:

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

The relation between the dissociation constant of acid and the base is given by the following equation-

$K_a=\frac{K_w}{K_b}$

Here, $K_w$ is ionic product of water with value ${10}^{-14}$.

Answer to Problem 1E

The concentration of $\left[{\text{OH}}^{-}\right]$ = $1.1\times {10}^{-13}\text{M}$.

Explanation of Solution

The chemical equation of this reaction is given below:

${\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}(aq){\text{+H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}^{\text{-}}(aq){\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}(aq)$

The concentration of $\left[{\text{OH}}^{-}\right]$ is calculated by the following equation:

$[{\text{OH}}^{-}]=\frac{K_w}{[{\text{H}}_{\text{3}}{\text{O}}^{+}]}$ ……………… (1)

Given that-

$\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]=0.089\text{2M}$

And,

$K_w=1\times {10}^{-14}$

Put the above values in equation (1)

$[{\text{OH}}^{-}]=\frac{1\times {10}^{-14}}{0.0892}\text{}\text{M}$

$[{\text{OH}}^{-}]=1.1\times {10}^{-13}\text{}\text{M}$

Interpretation Introduction

(c)

Interpretation:

A solution consists propionic acid, ${\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{COOH}$ of molarity 0.275 M and HI of molarity 0.0892 M. The concentration of $\left[{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{COO}}^{-}\right]$ is to be calculated. The value of equilibrium constant of acid is $K_a=1.3\times {10}^{-3}$.

Concept introduction:

The concentration of any solution is calculated by the molarity of the solution. It is defined as the number of moles of solute in 1 L of the solution.

$M=\frac{n}{V\left(\text{L}\right)}$

Here, n is number of moles of solute and V is volume of solution in L.

The dissociation constant of the weak acid is calculated by the following formula:

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

The expression can be written as follows:

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

Answer to Problem 1E

The concentration of $\left[{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{COO}}^{-}\right]$ = 4.0×10-5 M.

Explanation of Solution

The chemical equation of this reaction is given below:

${\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}(aq){\text{+H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}^{\text{-}}(aq){\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}(aq)$

Now, the concentration will be calculated by using the following ICE table:

	${\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}$	${\text{H}}_{\text{2}}\text{O}$	${\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_2^{-}$	${\text{H}}_{\text{3}}{\text{O}}^{+}$
Initial (I)	0.275	-	0	0.0892
Change (C)	-x	-	+x	+x
Equilibrium (E)	0.275-x	-	x	0.0892+x

From the ICE table:

${\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}^{\text{-}}{\text{ or CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}_2^{-}=x$

K_a = 1.3×10-5

${\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}$ = 0.275-x

${\text{H}}_{\text{3}}{\text{O}}^{+}$ = 0.0892+x

The equation for the K_a is given as

$K_a=\frac{{\text{[C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[C}}_{\text{3}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}\text{]}}$ …………… (1)

Put the given values in equation (1).

$1.3\times {10}^{-5}=\frac{x\left(0.0892+x\right)}{0.275-x}$

On calculation −

$x=4.0\times {10}^{-5}M$

Therefore, ${\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_2^{-}or{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}_2^{-}=x=4.0\times {10}^{-5}\text{M}$

Interpretation Introduction

(d)

Interpretation:

A solution consists propionic acid, ${\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{COOH}$ of molarity 0.275 M and HI of molarity 0.0892 M. The concentration of $\left[{\text{I}}^{-}\right]$ is to be calculated. The value of equilibrium constant of acid is $K_a=1.3\times {10}^{-3}$.

Concept introduction:

The dissociation constant of the weak acid is calculated by the following formula:

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

The expression can be written as follows:

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

The strong acid dissociates completely in comparison to the weak acid.

Answer to Problem 1E

The concentration of [HI] = 0.0892 M.

Explanation of Solution

The complete dissociation of strong acid occurs in the solution, whereas weak acid is not able to dissociate completely.

From the given question −

Hydroiodic acid, HI = strong acid

The molarity of [HI] or [H₃O⁺] ion = 0.0892 M

As hydroiodic acid is strong acid it will dissociate completely in the solution. Hence, the initial concentration of [HI] = 0.0892 M.