Consider an aqueous solution of HF. The molar heat of...

Consider an aqueous solution of HF. The molar heat of formation for aqueous HF is $-320.1 \mathrm{~kJ} / \mathrm{mol}$. (a) What is the $\mathrm{pH}$ of a $0.100 \mathrm{M}$ solution of $\mathrm{HF}$ at $100^{\circ} \mathrm{C} ?$ (b) Compare with the $\mathrm{pH}$ of a $0.100 \mathrm{M}$ solution of $\mathrm{HF}$ at $25^{\circ} \mathrm{C}$.

Consider an aqueous solution of HF. The molar heat of formation for aqueous HF is $-320.1 \mathrm{~kJ} / \mathrm{mol}$.
(a) What is the $\mathrm{pH}$ of a $0.100 \mathrm{M}$ solution of $\mathrm{HF}$ at $100^{\circ} \mathrm{C} ?$
(b) Compare with the $\mathrm{pH}$ of a $0.100 \mathrm{M}$ solution of $\mathrm{HF}$ at $25^{\circ} \mathrm{C}$.

Key Concepts

pH is a measure of the acidity of a solution, defined as the negative logarithm (base 10) of the hydrogen ion concentration. It is a key parameter in acid-base chemistry and helps describe the degree of ionization of acids and bases in solution.

Acid Dissociation Equilibrium

This concept refers to the reversible reaction in which a weak acid donates a proton (H+) to water to form its conjugate base and hydronium ion. The equilibrium position is described by the acid dissociation constant (Ka), which quantifies the extent to which the acid dissociates in solution.

Acid Dissociation Constant (Ka) and pKa

The acid dissociation constant, Ka, is a numerical measure of the strength of an acid in solution; a higher Ka indicates a stronger acid that dissociates more extensively. pKa, defined as the negative logarithm of Ka, provides a more convenient scale for comparing acid strengths. Both Ka and pKa are temperature dependent, meaning that changes in temperature can significantly impact the acid’s degree of ionization.

Temperature Dependence of Equilibrium

Temperature can affect the equilibrium position of chemical reactions including acid dissociation. Changes in temperature alter the equilibrium constant values (like Ka) according to thermodynamic principles such as Le Chatelier’s principle and the van'T Hoff equation. This results in a shift in the acid-base balance and consequently, a change in the pH of the solution.

van’t Hoff Equation

The van’t Hoff equation relates the change in the equilibrium constant to temperature changes. It is an important tool for understanding and predicting how the acidity (or basicity) of a solution will change with temperature, by accounting for the enthalpy change associated with the reaction.

Transcript

00:01 Okay, so we're trying to find the ka for the ionization of hf at 100 degrees celsius.

00:10 I can look it up at 25 degrees celsius.

00:13 In order to find it at a different temperature, i now need to go back to the van hoff equation, which is the ln of k2 over k1 equals delta h over r times 1 over t1, those are temperatures minus 1 over t2.

00:33 The equation we're looking for here is hf simply ionizing to give its ions, h plus, and f minus.

00:45 I'm going to need to find delta h for this.

00:48 So delta h of reaction, we can find it using heats of formation, so heats of formation of products minus reactants.

00:56 So the heat of formation of h plus is zero.

00:59 And the heat of formation of f minus is minus 332 .6.

01:05 Look them up in your tables.

01:07 And then i'll subtract the reactant, which is negative 320 .1.

01:13 So this will give us negative 12 .5 kilojoules.

01:19 Well, we know k .a.

01:21 For h .f.

01:23 At 25 degrees celsius.

01:26 You can look it up in the table, which is 6 .9 times 10 to negative 4.

01:31 So we want to find ka at 100 degrees celsius.

01:35 So let's start plugging in what we know.

01:36 So the ln of k2 over k1, which is going to be our 6 .9 times 10 and negative 4 is delta h.

01:50 Okay.

01:51 And since r is 8 .31 and it's in joules per mole kelvin, we're going to want delta h in joules.

02:00 So this will be negative 12 ,500, just multiplied by 1 ,000.

02:05 Okay.

02:06 And then one over our initial temperature was the 25 degrees celsius, so that'll be 298 kelvin.

02:13 And we're trying to find it at 100 degrees celsius, so that'll be 373 kelvin.

02:22 All right, so the ln of k2 minus the ln of 6 .9 times 10 and negative 4...

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