SCSA Chemistry Acids and bases

According to Arrhenius, acids are hydrogen-containing compounds that dissociate in water to give $\text{H}^+$ ions. $\text{HCl}(aq)$ would be considered an acid by Arrhenius as it produces $\text{H}^+$ ions in water.

Arrhenius would not recognise the salt $\text{NH}_4\text{Cl}$ as an acid, as the predominant ions present in aqueous solution are ammonium and chloride.

In Brønsted–Lowry theory, acids are defined as proton donors. $\text{HCl}(aq)$ is a proton donor and therefore a Br€™nsted–Lowry acid.

Ammonium chloride ($\text{NH}_4\text{Cl}$) is classified as a Br€™nsted–Lowry acid as the ammonium ion donates a proton to water and forms a hydronium ion.

$[\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-2.23} = 5.89 \times 10^{-3} \text{ mol L}^{-1}$

For 1.00 L solution, $n(\text{H}_3\text{O}^+) = n(\text{CH}_3(\text{CH})_4\text{COO}^-) = 5.89 \times 10^{-3} \text{ mol}$
For 1.00 L solution, $n(\text{CH}_3(\text{CH})_4\text{COOH}) = 0.250 \text{ mol}$

Weak acids undergo partial or incomplete ionisation in water. The answer to part (b) is numerically small, indicating that only a small percentage of sorbic acid in solution is ionised. The value of $K_a$ is less than one, which indicates a greater proportion of reactants compared to products.

The $K^+$ ions in solution are neutral and do not react with water, whereas the $HSO_3^-$ and $HCO_3^-$ ions undergo hydrolysis reactions. For the hydrolysis reactions for $HSO_3^-$, the reaction that produces $H_3O^+$ occurs to a greater extent than the reaction that produces $OH^-$, therefore the solution will be acidic. For the hydrolysis reactions for $HCO_3^-$, the reaction that produces $OH^-$ occurs to a greater extent than the reaction that produces $H_3O^+$, therefore the solution is basic. A basic solution has a greater concentration of $OH^-$ ions than $H_3O^+$ ions, and an acidic solution has a greater concentration of $H_3O^+$ ions than $OH^-$ ions.

The relevant equations for $HSO_3^-$ are (any 1 of the following):

• $HSO_3^-(aq) + H_2O(l) \rightleftharpoons SO_3^{2-}(aq) + H_3O^+(aq)$
• $HSO_3^-(aq) + H_2O(l) \rightleftharpoons H_2SO_3(aq) + OH^-(aq)$.

The relevant equations for $HCO_3^-$ are (any 1 of the following):

• $HCO_3^-(aq) + H_2O(l) \rightleftharpoons H_2CO_3(aq) + OH^-(aq)$
• $HCO_3^-(aq) + H_2O(l) \rightleftharpoons CO_3^{2-}(aq) + H_3O^+(aq)$.

Carbon
$n(C) = n(CO_2) = 1.320/44.01 = 0.02999$ mol
$m(C) = 0.02299 \times 12.01 = 0.3602$ g

Chlorine
$n(Cl) = n(HCl) = n(NH_3) = 1.050 \times 0.08570 = 0.08999$ mol
$m(Cl) = 0.08999 \times 35.45 = 3.1900$ g

Fluorine
$m(F) = 4.121 - (0.3602 + 3.1900) = 0.5708$ g
$n(F) = 0.5708 / 19.00 = 0.03004$ mol

Mole Ratio
C: 0.02999
Cl: 0.08999
F: 0.03004

Simplify
Divide by 0.02999
C: 1
Cl: 3.00
F: 1.002

Empirical Formula
$CCl_3F$

Molecular Formula
Empirical formula mass (EFM) = 137.36 amu/g mol$^{-1}$
$n = PV/RT = (120.0 \times 0.6068)/(8.314 \times 323.75) = 0.027052476$ mol
Molecular formula mass (MFM) = $m/n = 3.721/0.0276503204 = 137.54$ g mol$^{-1}$
Molecular formula = MFM/EFM $\times$ Empirical formula = $137.54/137.36 \times CCl_3F = 1.0013 \times CCl_3F$

Molecular formula = Empirical formula = $CCl_3F$

Equation:

NH$_4^+$ is labelled as an acid and NH$_3$ is labelled as a base.

The strong acid produced in the experiment increases the hydrogen ion concentration. The buffer equilibrium will shift to the left to counteract the imposed change, as H$_3$O$^+$ reacts with NH$_3$, and the pH will only drop slightly.

Buffer capacity represents the amount of acid or base that can be added before the pH changes by more than one unit of pH.

• The actual concentration of both the weak acid and its conjugate base.
• The relative concentrations of the weak acid and its conjugate base.

$CH_3COOH(aq) + OH^-(aq) \rightleftharpoons CH_3COO^-(aq) + H_2O(l)$

The conjugate base formed is $CH_3COO^-$.

$K_b = \frac{1.00 \times 10^{-14}}{10^{-4.76}} = 10^{-9.24} = 5.75 \times 10^{-10}$

At equivalence point, moles $OH^-$ = moles $H^+$ = moles $CH_3COO^-$
moles $CH_3COO^- = n \times V = 0.10 \times 0.015 = 1.50 \times 10^{-3}$

$[CH_3COOH] = [OH^-] = x$

$K_b = \frac{[OH^-][CH_3COOH]}{[CH_3COO^-]}$

$K_b = 5.75 \times 10^{-10} = \frac{x^2}{[2.31 \times 10^{-2}]}$

$x^2 = 1.33 \times 10^{-11}$

$x = 3.64 \times 10^{-6} = [OH^-]$

$pOH = -\log(3.64 \times 10^{-6}) = 5.44 \sim 5.4$

$pH = 14 - 5.4 = 8.6$