Name: salt hydrolysis and ph in solutions
Uploaded: 2020-09-24T15:06:51.000Z
Duration: 4 min 8 s
Channel: JoVE

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Determining the pH of Salt Solutions

Les solutions salines peuvent être acidiques, basiques ou neutres. Les ions individuels peuvent être examinés pour déterminer si le pH de sa solution sera inférieur, supérieur, ou égal à sept. Sels contenant des cations et des anions au pH neutre forment des solutions neutres.Ces sels sont composés des contre-ions d'un acide fort et une base forte. Une solution aqueuse pure de nitrate de potassium aura un pH de 7, car ni potassium ni nitrate possède des propriétés acidiques ou basiques. Les sels contenant un cation ce qui est le contre-ion d'une base faible et d'un anion ce qui est le contre-ion d'un acide fort auront un pH inférieur à 7.Dans une solution de bromure d'ammonium, les ions bromure ont un pH neutre, tandis que les ions ammonium sont acidiques et donnent des protons aux molécules d'eau. Les sels contenant un cation ce qui est le contre-ion d'une base forte et d'un anion ce qui est le contre-ion d'un acide faible auront un pH supérieur à 7. Dans une solution d'acétate de sodium, les ions sodium ont un pH neutre, tandis que les ions acétate sont basiques et acceptent les protons des molécules d'eau.Certains sels contiennent un cation et un anion qui sont respectivement des contre-ions d'un acide faible et d'une base faible. La solution sera acidique si le Ka du cation est plus grand que le Kb de l'anion;basique si le Kb de l'anion est plus grand que le Ka du cation;ou neutre si le Ka et le Kb sont égaux. Dans une solution de nitrite d'ammonium, le Ka pour l'ammonium est supérieur au Kb pour les ions nitrite;par conséquent, la solution sera acidique.Le pH exact d'une solution saline contenant un ion acidique ou basique peut être déterminé utilisant la constante de dissociation de l'acide conjugué ou base de cet ion. Le pH du cyanure de sodium de 0, 35 molaire peut être déterminé en calculant le Kb de cyanure à partir du Ka de son acide conjugué acide cyanhydrique et mise en place d'une table ICE pour déterminer la concentration d'équilibre. Les cations sodium ne réagissant pas avec l'eau, ils n'affectent pas le pH et peuvent être omis.En revanche, les ions cyanure acceptent les protons des molécules d'eau et générent des ions hydroxyde. Le Kb pour les ions cyanure peut être déterminé en utilisant l'équation Ka fois Kb est égal à Kw.Comme le Ka de l'acide cyanhydrique est de 4, 90 10⁻¹⁰, le Kb pour les ions cyanure est de 2, 04 10⁻⁵. Le Kb est égal à la concentration d'acide cyanhydrique multiplié par la concentration d'ions hydroxyde divisé par la concentration d'ions cyanure.Une table ICE peut être construite pour exprimer les concentrations initiale et d'équilibre. En substituant les concentrations d'équilibre dans l'expression, Kb est égal à x fois x divisé par 0, 35, qui peut être vérifié ultérieurement par la règle des 5%En résolvant l'équation, x est égal à 2, 7 10⁻³ molaire. Comme x n'est que 0, 77%de 0, 35 M, l'approximation 0, 35 moins x est égal à 0, 35 est valide.Le pOH de la solution peut être calculé en prenant le log négatif de 2, 7 10⁻³ M, qui est égal à 2, 57. Le pH de la solution est déterminé en utilisant l'équation pH plus pOH est égal à 14;par conséquent, le pH de cette solution de cyanure de sodium est de 11, 43.

détermination du ph des solutions salines

Le pH d'une solution de sels est déterminé grâce à ses composants anions et cations. Les sels qui contiennent des anions de pH-neutre et des cations produisant les ions hydronium forment une solution dont le pH est inférieur à 7. Par exemple, dans une solution de nitrate d'ammonium (NH4NO3), les ions NO3− ne réagissent pas avec l'eau alors que les ions NH4+ produisent les ions hydronium qui rendent la solution acide.  En revanche, les sels qui contiennent des cations de pH-neutre et des anions produisant les ions hydroxyde forment une solution dont le pH est supérieur à 7. Par exemple, dans une solution de fluorure de sodium (NaF), le Na+ est de pH neutre, mais le F- produit les ions hydroxyde et forme la solution basique. Les contre-ions d'un acide fort ou d'une base forte sont de pH-neutre et les sels formés par de tels contre-ions forment une solution neutre avec un pH égal à 7. Par exemple, dans KBr, le cation K+ est inerte et n'influence pas le pH. L'ion bromure est la base conjuguée d'un acide fort, et la force de sa base est donc négligeable (pas d'ionisation de la base notable). La solution est neutre. 
 Certains sels contiennent à la fois un cation acide et un anion basique. L'acidité ou la basicité globale d'une solution est déterminée par la force relative du cation et de l'anion, qui peuvent être comparées à l'aide de Ka et de Kb. Par exemple, dans NH4F, l'ion NH4+ est acide et l'ion F− est basique (base conjuguée de l'acide faible HF). En comparant les deux constantes d'ionisation : Ka de NH4+ est de 5,6 × 10−10 et le Kb de F− est de 1,6 × 10−11, de sorte que la solution est acide, puisque Ka > Kb. 
 Calcul du pH d'une solution de sel acide 
 L'aniline est une amine utilisée pour fabriquer des colorants. Elle est isolée sous forme de chlorure d'anilinium [C6H5NH3+]Cl, un sel préparé par la réaction de la base faible aniline et de l'acide chlorhydrique. Quel est le pH d'une solution 0,233 M de chlorure d'anilinium ? 
 <img alt="Eq1" src="/files/ftp_upload/11404/11404_Equation_1.png" style="height: 50px;" />
 Le Ka de l'ion anilinium est déduit à partir du Kb de sa base conjuguée, l'aniline: 
 <img alt="Eq2" src="/files/ftp_upload/11404/11404_Equation_2.png" style="height: 90px;" />
 À l'aide des informations fournies, un tableau ICE est préparé pour ce système : 
 <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;">C6H5NH3+ (aq) </td>
 <td style="color: black; border: 1px solid black; text-align: center;">H3O+ (aq) </td>
 <td style="color: black; border: 1px solid black; text-align: center;">C6H5NH2 (aq) </td>
 </tr>
 <tr>
 <td style="color: black; border: 1px solid black; text-align: center;">Concentration initiale (M) </td>
 <td style="color: black; border: 1px solid black; text-align: center;">0,233 </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;">Changement (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;">Concentration à l'équilibre (M) </td>
 <td style="color: black; border: 1px solid black; text-align: center;">0,233 − 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>
 </tbody>
 </table>
 Remplacer ces termes de concentration à l'équilibre dans l'expression de Ka donne 
 <img alt="Eq3" src="/files/ftp_upload/11404/11404_Equation_3.png" style="height: 90px;" />
 En supposant que x << 0,233, l'équation est simplifiée et résolue pour trouver x :  
 <img alt="Eq4" src="/files/ftp_upload/11404/11404_Equation_4.png" style="height: 90px;" />
 <img alt="Eq5" src="/files/ftp_upload/11404/11404_Equation_5.png" style="height: 50px;" />
 Le tableau ICE définit x comme étant la molarité en ions hydronium, et le pH est donc calculé comme 
 <img alt="Eq6" src="/files/ftp_upload/11404/11404_Equation_6.png" style="height: 50px;" />
 Hydrolyse de [Al(H2O)6]3+ 
 Calculer le pH d'une solution de chlorure d'aluminium de 0,10 M qui se dissout complètement pour donner l'ion d'aluminium hydraté [Al(H2O)6]3+ en solution. 
 L'équation de la réaction et de Ka sont les suivantes : 
 <img alt="Eq7" src="/files/ftp_upload/11404/11404_Equation_7.png" style="height: 90px;" />
 Un tableau ICE avec les informations fournies est 
 <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;">Al(H2O)63+ (aq) </td>
 <td style="color: black; border: 1px solid black; text-align: center;">H3O+ (aq) </td>
 <td style="color: black; border: 1px solid black; text-align: center;">Al(H2O)5(OH)2+ (aq) </td>
 </tr>
 <tr>
 <td style="color: black; border: 1px solid black; text-align: center;">Concentration initiale (M) </td>
 <td style="color: black; border: 1px solid black; text-align: center;">0,10 </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;">Modifier (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;">Concentration à l'équilibre (M) </td>
 <td style="color: black; border: 1px solid black; text-align: center;">0,10 − 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>
 </tbody>
 </table>
 Remplacer les expressions des concentrations à l'équilibre dans l'équation de la constante d'ionisation donne : 
 <img alt="Eq8" src="/files/ftp_upload/11404/11404_Equation_8.png" style="height: 90px;" />
 En supposant que x << 0,10 et en résolvant l'équation simplifiée, vous obtenez : 
 <img alt="Eq9" src="/files/ftp_upload/11404/11404_Equation_9.png" style="height: 50px;" />
 Le tableau ICE définit x comme étant égal à la concentration en ions hydronium, le pH calculé est donc à 2,92 et la solution est acide. 
 <img alt="Eq10" src="/files/ftp_upload/11404/11404_Equation_10.png" style="height: 90px;" />
 Ce texte est adapté de<a href="https://openstax.org/books/chemistry-2e/pages/14-4-hydrolysis-of-salts"> Openstax, Chimie 2e, Section 14.4 : Hydrolyse des sels</a>.

Détermination du pH des solutions salines

Le pH d’une solution saline est déterminé par ses anions et ses cations constitutifs. Les sels qui contiennent des anions au pH neutre et les cations producteurs d’ions hydronium ...

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salt hydrolysis and ph in solutions

Salt hydrolysis and pH in solutions depend on the ions released when a salt dissolves. Some ions react with water, while others do not. The balance between the cation and anion helps show whether the solution becomes acidic, basic, or neutral.

Salts with pH-neutral anions and cations that produce hydronium ions give acidic solutions with pH below 7. Ammonium nitrate is one example. In NH4NO3, NO3− does not react with water, but NH4+ forms hydronium ions, so the solution is acidic.

Other salts contain pH-neutral cations and anions that produce hydroxide ions. These salts form basic solutions with pH above 7. Sodium fluoride is an example. In NaF, Na+ is pH-neutral, while F− reacts with water to produce hydroxide ions.

Salts made from the counterions of a strong acid or a strong base are usually neutral. Their solutions have a pH of 7. In KBr, K+ is inert and does not change the pH. Br− is the conjugate base of a strong acid, so it has negligible base strength and does not appreciably ionize in water.

Some salts contain both an acidic cation and a basic anion. In those cases, the overall pH depends on which ion is stronger. The ionization constants Ka and Kb are used to compare them. For NH4F, NH4+ is acidic and F− is basic. Because Ka for NH4+ is 5.6 × 10−10 and Kb for F− is 1.6 × 10−11, the solution is acidic since Ka is greater than Kb.

The pH of an acidic salt solution can be calculated with an ICE table. Anilinium chloride, [C6H5NH3+]Cl, comes from the weak base aniline and hydrochloric acid. For a 0.233 M solution, the Ka value for the anilinium ion is found from the Kb of aniline. The ICE table gives x as the hydronium ion concentration, and the pH is then calculated from that value.

Hydrolysis of the hydrated aluminum ion works the same way. Aluminum chloride dissolves completely to give [Al(H2O)6]3+ in solution. For a 0.10 M solution, the reaction and Ka are set up in an ICE table. After assuming x is much smaller than 0.10 and solving the simplified equation, the hydronium ion concentration gives a pH of 2.92. The solution is acidic.

Salt Hydrolysis and pH in Solutions

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Chemical bonding theories were pioneered by American chemist Gilbert&#xA0;N. Lewis. He developed a model called the Lewis model to explain the type and ...

Ionic Bonding and Electron Transfer

Ions are atoms or molecules bearing an electrical charge. A cation (a positive ion) forms when a neutral atom loses one or more electrons from its valence ...

Covalent Bonding and Lewis Structures

Compared to ionic bonds, which results from the transfer of electrons between metallic and nonmetallic atoms, covalent bonds result from the mutual ...

Bonding in Metals

Metallic bonds are formed between two metal atoms. A simplified model to describe metallic bonding has been developed by Paul Dr&#xFC;de called the ...

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; ...

IUPAC Nomenclature of Aldehydes

Aldehydes are named based on the systematic nomenclature rules set by the IUPAC. For acyclic aldehydes, the longest carbon chain containing the aldehydic ...

IUPAC Nomenclature of Ketones

Like aldehydes, ketones are named using IUPAC rules; in this case, by replacing &#8220;e&#8221; in the name of the longest hydrocarbon chain with ...

IUPAC Nomenclature of Carboxylic Acids

IUPAC names of carboxylic acids are systematically derived following a few rules discussed below.
For acyclic saturated monocarboxylic acids, the longest ...

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.

Salt Hydrolysis and pH in Solutions

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