Name: weak base ph and kb calculations
Uploaded: 2020-09-24T15:06:51.000Z
Duration: 3 min 21 s
Channel: JoVE

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Physical Science

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Weak Base Solutions

Una base debole, come l'ammoniaca, è una base di Brønsted che accetta un protone dall'acqua per produrre lo ione idrossido. Le basi deboli reagiscono parzialmente con l'acqua secondo la loro costante di dissociazione delle basi, Kb, che per l'ammoniaca è 1, 76 10⁻⁵. Il valore della Kb per l'ammoniaca può essere espresso come la concentrazione di ioni ammonio moltiplicata per la concentrazione di ioni idrossido divisa per la concentrazione di ammoniaca in equilibrio.

Kb può essere utilizzata per determinare la concentrazione di ioni idrossido in una soluzione di una base debole e, di conseguenza, il pOH e il pH della soluzione. La concentrazione di ioni idrossido e il pH di 0, 23 M della soluzione di ammoniaca possono essere determinati usando la sua costante di dissociazione delle basi e preparando una tabella ICE, contenente i valori iniziali e di equilibrio di ammoniaca, ioni ammonio e ioni idrossido. Sostituendo le concentrazioni di equilibrio nell'espressione per la Kb, Kb è uguale a x volte x diviso per 0, 23 meno x.

Poiché le basi deboli mostrano una dissociazione parziale, 0, 23 meno x può essere considerato approssimativamente 0, 23. Quando l'equazione è risolta, x è uguale a 2 10⁻³ molare. L'approssimazione 0, 23 meno x è uguale a 0, 23, è valida qui, poiché la concentrazione di ioni idrossido è solo lo 0, 86 percento di 0, 23 molare.

Per calcolare il pH di questa soluzione, occorre determinare prima il pOH, prendendo il log negativo della concentrazione di ioni idrossido, che è uguale a 2, 70. Il pH può essere determinato utilizzando la formula:pH più pOH pari a 14 e calcolato essere 11, 30. Il valore della Kb per una soluzione di una base debole può essere calcolato se il pH della soluzione della base debole è noto.

La metilammina è una base debole che si dissocia parzialmente in acqua in ioni metilammonio e ioni idrossido. Il valore della Kb per la metilammina può essere espresso come la concentrazione di ioni metilammonio, moltiplicata per la concentrazione di ioni idrossido divisa per la concentrazione di metilammina in equilibrio. Per calcolare la Kb di una soluzione di metilammina molare 0, 040 a pH 11, 6, è necessario prima calcolare il pOH seguito dalla sua concentrazione di ioni idrossido.

Poiché il pH è 11, 60, il pOH è 2, 40 e la sua concentrazione di ioni idrossido è 4, 0 10³. La tabella ICE può essere costruita dalle concentrazioni iniziali e di equilibrio di metilammina, metilammonio e idrossido. Sostituendo le concentrazioni in equilibrio nell'espressione per Kb si ottiene il valore della Kb, che è 4, 4 10⁻⁴.

soluzioni a base debole

Alcuni composti producono ioni idrossido quando disciolti reagendo chimicamente con le molecole d'acqua. In tutti i casi questi composti reagiscono solo parzialmente e quindi sono classificati come basi deboli. Questi tipi di composti sono anche abbondanti in natura e importanti materie prime in varie tecnologie. Ad esempio, la produzione globale della base debole ammoniaca è in genere ben superiore alle 100 tonnellate all’anno, essendo ampiamente utilizzata come fertilizzante agricolo, materia prima per la sintesi chimica di altri composti e ingrediente attivo nei detergenti domestici. Quando disciolta in acqua, l'ammoniaca reagisce parzialmente per produrre ioni idrossido, come mostrato qui:

<img alt="Eq1" src="/files/ftp_upload/11401/11401_Equation_1.png" style="height: 50px;" />

Questa è, per definizione, una reazione acido-base, in questo caso comporta il trasferimento di ioni H^+ dalle molecole d'acqua alle molecole di ammoniaca. In condizioni tipiche, solo circa l’1% dell’ammoniaca disciolta è presente come ioni NH4^+.

Calcolo delle concentrazioni di ioni idrossido e pOH in una soluzione a base debole

Trova la concentrazione dello ione idrossido, il pOH e il pH di una soluzione 0,25 M di trimetilammina, una base debole:

<img alt="Eq2" src="/files/ftp_upload/11401/11401_Equation_2.png" style="height: 50px;" />

La tabella ICE per questo sistema è

<table style="margin-left: auto; margin-right: auto;">
	<tbody>
		<tr>
			<td style="color: black; border: 1px solid black; text-align: center;"></td>
			<td style="color: black; border: 1px solid black; text-align: center;">(CH3)3N (aq)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">(CH3)3NH+ (aq) </td>
			<td style="color: black; border: 1px solid black; text-align: center;">OH− (aq) </td>
		</tr>
		<tr>
			<td style="color: black; border: 1px solid black; text-align: center;">Initial Concentration (M)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0.25</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0</td>
			<td style="color: black; border: 1px solid black; text-align: center;">~0</td>
		</tr>
		<tr>
			<td style="color: black; border: 1px solid black; text-align: center;">Change (M)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">−x</td>
			<td style="color: black; border: 1px solid black; text-align: center;">+x</td>
			<td style="color: black; border: 1px solid black; text-align: center;">+x</td>
		</tr>
		<tr>
			<td style="color: black; border: 1px solid black; text-align: center;">Equilibrium Concentration (M)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0.25 − x</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0 + x</td>
			<td style="color: black; border: 1px solid black; text-align: center;">~0 + x</td>
		</tr>
	</tbody>
</table>

Sostituendo i termini di concentrazione all'equilibrio nell'espressione K_b si ottiene

<img alt="Eq3" src="/files/ftp_upload/11401/11401_Equation_3.png" style="height: 90px;" />

Supponendo x << 0,25 e risolvendo per x si ottiene

<img alt="Eq4" src="/files/ftp_upload/11401/11401_Equation_4.png" style="height: 50px;" />

Questo valore è inferiore al 5% della concentrazione iniziale (0,25), quindi l'ipotesi è giustificata.

Come definito nella tabella ICE, x è uguale alla concentrazione di equilibrio dello ione idrossido:

<img alt="Eq5" src="/files/ftp_upload/11401/11401_Equation_5.png" style="height: 50px;" />

Si calcola che il pOH sia

<img alt="Eq6" src="/files/ftp_upload/11401/11401_Equation_6.png" style="height: 50px;" />

Utilizzando la relazione;

<img alt="Eq7" src="/files/ftp_upload/11401/11401_Equation_7.png" style="height: 50px;" />

permette il calcolo del pH:

<img alt="Eq8" src="/files/ftp_upload/11401/11401_Equation_8.png" style="height: 50px;" />

Determinazione della K_b dal pH

Se il pH di una soluzione 0,28 M di etilammina (C_2H_5NH_2) è 12,10, qual è il suo K_b?

<img alt="Eq9" src="/files/ftp_upload/11401/11401_Equation_9.png" style="height: 50px;" />

Per calcolare la K_b dell'etilammina è necessario determinare prima il pOH e la concentrazione degli ioni idrossido. Poiché il pH è 12,10, il pOH può essere calcolato come segue:

<img alt="Eq10" src="/files/ftp_upload/11401/11401_Equation_10.png" style="height: 50px;" />

Poiché il pOH è 1,90, la concentrazione di ioni idrossido della soluzione può essere calcolata utilizzando la formula

<img alt="Eq11" src="/files/ftp_upload/11401/11401_Equation_11.png" style="height: 50px;" />

La tabella ICE può essere costruita per questo sistema come segue

<table style="margin-left: auto; margin-right: auto;">
	<tbody>
		<tr>
			<td style="color: black; border: 1px solid black; text-align: center;"></td>
			<td style="color: black; border: 1px solid black; text-align: center;">C2H5NH2 (aq)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">C2H5NH3+ (aq)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">OH− (aq)</td>
		</tr>
		<tr>
			<td style="color: black; border: 1px solid black; text-align: center;">Initial Concentration (M)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0.28</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0</td>
			<td style="color: black; border: 1px solid black; text-align: center;">~0</td>
		</tr>
		<tr>
			<td style="color: black; border: 1px solid black; text-align: center;">Change (M)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">−0.0126</td>
			<td style="color: black; border: 1px solid black; text-align: center;">+0.0126</td>
			<td style="color: black; border: 1px solid black; text-align: center;">+0.0126</td>
		</tr>
		<tr>
			<td style="color: black; border: 1px solid black; text-align: center;">Equilibrium Concentration (M)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0.28 − 0.0126</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0.0126</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0.0126</td>
		</tr>
	</tbody>
</table>

Poiché 0,0126 M è il 4,5% di 0,28 M, 0,28 − 0,0126 può essere considerato quasi uguale a 0,28 M secondo la regola del 5%.

Dopo aver sostituito i valori sopra riportati nell'espressione per la K_b di etilammina,

<img alt="Eq12" src="/files/ftp_upload/11401/11401_Equation_12.png" style="height: 90px;" />

Questo testo è adattato da Openstax, Chemistry 2e <a href="https://openstax.org/books/chemistry-2e/pages/4-2-classifying-chemical-reactions">Section 4.2: </a><a href="https://openstax.org/books/chemistry-2e/pages/4-2-classifying-chemical-reactions">Classifying Chemical Reactions</a> and <a href="https://openstax.org/books/chemistry-2e/pages/14-3-relative-strengths-of-acids-and-bases">14.3 Relative Strengths of Acids and Bases.</a>

Soluzioni a base debole

Alcuni composti producono ioni idrossido quando disciolti reagendo chimicamente con le molecole d'acqua. In tutti i casi, questi composti reagiscono solo parzialmente e ...

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weak base ph and kb calculations

Weak base pH and Kb calculations show how partially reacting bases change water chemistry. Weak bases produce hydroxide ions when they dissolve by reacting with water molecules. Because the reaction only happens partly, these compounds are classified as weak bases.

Weak bases are common in nature and in technology. Ammonia is one major example. Global production of ammonia is typically well over 100 metric tons each year. It is used as an agricultural fertilizer, a starting material for chemical synthesis, and an ingredient in household cleaners.

When ammonia dissolves in water, it reacts only partially to form hydroxide ions. This is an acid-base reaction because H+ ions move from water molecules to ammonia molecules. Under typical conditions, only about 1% of the dissolved ammonia is present as NH4+ ions.

The video also shows how to calculate hydroxide ion concentration, pOH, and pH for a weak base solution. For a 0.25 M trimethylamine solution, an ICE table is used to track initial, change, and equilibrium concentrations. The Kb expression is then written with the equilibrium terms, and the x &lt;&lt; 0.25 assumption is checked with the 5% rule before solving for x as the hydroxide ion concentration. From there, pOH is found and pH is calculated.

A second example shows how to determine Kb from pH. For a 0.28 M ethylamine solution with a pH of 12.10, the pOH is first found, then the hydroxide ion concentration is calculated. An ICE table is built for ethylamine, ethylammonium ion, and hydroxide ion. After applying the 5% rule, the values are substituted into the Kb expression to find the base dissociation constant.

Weak Base pH and Kb Calculations

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weak base solutions

Weak base pH and Kb calculations show how partially reacting bases change water chemistry. Weak bases produce hydroxide ions when they dissolve by reacting with water molecules. Because the reaction only happens partly, these compounds are classified as weak bases. Weak bases are common in nature and in technology. Ammonia is one major example. Global production of ammonia is typically well over 100 metric tons each year. It is used as an agricultural fertilizer, a starting material for...

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key terms and concepts

Learn how weak bases form hydroxide ions, then use ICE tables to find pOH, pH, and Kb for trimethylamine and ethylamine examples.

Weak Base pH and Kb Calculations | JoVE K12

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Ohio’s Learning Standards and Model Curriculum

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New York State Science Learning Standards

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Classifying Matter by State

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Classifying Matter by Composition

Matter: Pure Substances and Mixtures
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Atoms and Molecules

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Colloids

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Construct an explanation to classify matter as a pure substance or mixture; homogeneous or heterogeneous; element or compound; solution, colloid or suspension.

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Phase Transitions

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Phase Transitions: Vaporization and Condensation

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Vapor Pressure

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Clausius-Clapeyron Equation

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Phase Transitions: Melting and Freezing

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Phase Transitions: Sublimation and Deposition

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Use models to compare the phases of matter and the physical changes they undergo.

PS.PSc.1.2

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Physical and Chemical Properties of Matter

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Physical Properties of Matter

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The Atomic Theory of Matter

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Subatomic Particles

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Use models to interpret the data presented in Bohr diagrams and electron dot diagrams for neutral atoms of elements 1 through 18.

PS.PSc.1.4

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Use models to compare representations of atoms, ions, and isotopes.

PS.PSc.1.5

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Noble Gases


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Use the Periodic Table as a model to predict the relative properties (metallic/nonmetallic character, ionic charge, and reactivity) and arrangement of elements based on the pattern of valence electrons in the outermost energy levels of atoms.

PS.PSc.1.6

Types of Chemical Bonds

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Ionic Bonding and Electron Transfer

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Covalent Bonding and Lewis Structures

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Bonding in Metals

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Construct an explanation to classify the type of chemical bond that occurs (covalent, ionic, or metallic) in a given substance.

PS.PSc.2.1

Nomenclature of Alkanes

In the late 19th-century, the number of new chemical compounds discovered increased tremendously. Hence, the necessity arose to develop a naming system ...

Nomenclature of Alkenes

The IUPAC naming system for alkenes replaces -an- with -en- in the corresponding parent alkanes. Accordingly, a simple alkene replaces the -ane suffix of ...

Nomenclature of Alkynes

Alkynes are unsaturated hydrocarbons characterized by the presence of carbon-carbon triple bonds and have a general formula CnH2n-2. The nomenclature of ...

structure-and-nomenclature-of-alcohols-and-phenols

Structure and Nomenclature of Alcohols and Phenols

Overview 
Alcohols are one of the most important functional groups in organic chemistry. The name of alcohol comes from the hydrocarbon from which it is ...

Structure and Nomenclature of Ethers

Structure and Bonding
Ethers are organic compounds with an ether functional group which is characterized by an oxygen atom connected to two &#8212; ...

Use models to apply International Union of Pure and Applied Chemistry (IUPAC) conventions to name and write formulas for simple compounds.

PS.PSc.2.2

Understand types, properties, and structure of matter.

Weak Base pH and Kb Calculations

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