Chapter 8, Problem 105P

Interpretation Introduction

(a)

Interpretation:

The molarity of the hydrofluoric acid in the solution is to be calculated.

Concept Introduction:

The molarity of a solution is defined as number of moles of solute in 1L of solution. This is mathematically represented as follows:

$M=\frac{n}{V(in\text{l})}$

Here, n is number of moles and v is volume of solution.

The pressure, volume, moles and temperature of a gas are related to each other according to an equation,

$PV=nRT$

Here, p is pressure, v is volume, n is number of moles, r is gas constant and t is temperature of the gas.

Answer to Problem 105P

Thus, the molarity of the HF solution is 0.02498 M.

Explanation of Solution

The pressure, volume, moles and temperature of a gas are related to each other according to an equation,

$PV=nRT$

Here,

$P$ = Pressure of HF gas = 0.601 atm.

$V$ = Volume of HF gas = 1.00 L.

$n$ = Moles of HF gas = it is to be found.

$R$ = Gas constant = 0.0821 L.atm/mol.K.

$T$ = Temperature in K = 20.0 $\text{C}$ = 20.0 + 273 = 293.0 K.

Putting the given values in the ideal gas equation, we get.

$\begin{array}{l}PV=nRT\\ (0.601\cancel{\text{atm}})(1.00\cancel{\text{L}})=n(0.0821\frac{\cancel{\text{L}}\text{.}\cancel{\text{atm}}}{\text{mol}\text{.}\cancel{\text{K}}})(293.0\cancel{\text{K}})\\ n=\frac{0.601\times 1.00}{0.0821\times 293.0}\\ n=0.02498\text{ mol}\end{array}$

Thus, the number of moles of HF = 0.02498 mol.

Now, the molarity of this solution is to be calculated.

$\begin{array}{c}M=\frac{n}{V(in\text{l})}\\ =\frac{0.02498\text{mol}}{1.00\text{L}}\\ =0.02498\text{ M}\end{array}$

Thus, the molarity of the HF solution is 0.02498 M.

Interpretation Introduction

(b)

Interpretation:

The acid ionization constant, $K_{\text{a}}$ is to be calculated.

Concept Introduction:

Weak acids do not dissociate completely. Let HA be a weak acid. The dissociation of the weak acid can be represented by the chemical equation,

$\text{HA}(\text{a}q)\underset{}{\rightleftharpoons }{\text{H}}^{+}(\text{a}q)+{\text{A}}^{-}(\text{a}q)$

The equation for acid dissociation constant can be written from this chemical equation.

$K_{\text{a}}=\frac{[{\text{H}}^{+}][{\text{A}}^{-}]}{[\text{HA}]}$.

Answer to Problem 105P

The acid dissociation constant of hydrofluoric acid is $K_{\text{a}}=0.01472$.

Explanation of Solution

Hydrofluoric acid is a weak acid. Hence, it does not dissociate completely. The dissociation of the given weak acid can be represented by the chemical equation,

$\text{HF}(\text{a}q)\underset{}{\rightleftharpoons }{\text{H}}^{+}(\text{a}q)+{\text{F}}^{-}(\text{a}q)$

The equation for acid dissociation constant can be written from this chemical equation.

$K_{\text{a}}=\frac{[{\text{H}}^{+}][{\text{F}}^{-}]}{[\text{HF}]}$

The concentrations of each of the ions at equilibrium can be obtained from the ICE table, where ICE represents the Initial, Change and Equilibrium concentrations of the weak acid.

$\begin{array}{l}\begin{array}{l}\text{ HF}(\text{a}q)\underset{}{\rightleftharpoons }{\text{H}}^{+}(\text{a}q)+{\text{F}}^{-}(\text{a}q)\\ {\text{ [HF] [H}}^{+}][{\text{F}}^{-}]\\ \text{i} 0.02498 0 0\end{array}\hfill \\ \hfill \\ \text{C} -\text{X} +\text{X} +\text{X}\hfill \\ \hfill \\ \text{e} 0.02498-\text{X} +\text{X} +\text{X}\hfill \end{array}$

The hydrogen ion concentration can be obtained from the given pH. The pH is defined as the negative logarithm of the hydrogen ion concentration.

$p\text{H}=-\log [{\text{H}}^{+}]$

The pH of the weak acid solution at equilibrium is 1.88. Thus, we can calculate the concentration of the hydrogen ion.

$\begin{array}{l}p\text{H}=-\log [{\text{H}}^{+}]\\ [{\text{H}}^{+}]={10}^{-1.88}=0.01318\text{mol}{\text{L}}^{-1}\\ \text{X}=[{\text{H}}^{+}]=0.01318\text{mol}{\text{L}}^{-1}\end{array}$

We calculated the “x” which is the concentration of hydrogen ion. The concentration of the anion is also “x”. Thus,

$\text{X}=[{\text{F}}^{-}]=0.01318\text{mol}{\text{L}}^{-1}$

Now, we need to calculate the concentration of $[\text{H}\text{F}]$

$\begin{array}{c}[\text{HF}]=0.02498-\text{X}\\ =0.02498-0.01318\\ =0.0118M\end{array}$

Thus, the concentration of hydrofluoric acid is $[\text{H}\text{F}]$ = 0.0118.

The concentrations of the anion, hydrogen ion and hydrofluoric acid are used in the equation used for acid dissociation constant.

$\begin{array}{c}{K}_{\text{a}}=\frac{[{\text{H}}^{+}][{\text{F}}^{-}]}{[\text{HF}]}\\ =\frac{[0.01318][0.01318]}{0.0118}\\ K_{\text{a}}=0.01472\end{array}$

Thus, the acid dissociation constant of hydrofluoric acid is $K_{\text{a}}=0.01472$.