Name: buffer ph after acid or base addition
Uploaded: 2020-09-24T15:06:51.000Z
Duration: 2 min 45 s
Channel: JoVE

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Calculating pH Changes in a Buffer Solution

Nieznaczna zmiana pH zachodzi w roztworze buforowym po dodaniu niewielkiej ilości mocnego kwasu lub dodaniu niewielkiej ilości mocnej zasady.
Ta zmiana pH jest obliczana w dwóch odrębnych krokach.
Po pierwsze, obliczenia stechiometryczne służą do określenia zmiany stężeń.
Następnie stosuje się obliczenia równowagi w celu określenia nowego pH roztworu, korzystając z tabeli ICE lub równania Hendersona-Hasselbalcha. 
pH buforu zawierającego 2,0 M zarówno kwasu fluorowodorowego, jak i fluorku sodu można obliczyć przed i po dodaniu mocnego kwasu lub zasady.
Równanie Hendersona-Hasselbalcha można wykorzystać do określenia początkowego pH tego buforu, ponieważ stężenia są wysokie w stosunku do Ka słabego kwasu, a zmiana stężeń jest mniejsza niż 5%.
P Ka dla kwasu fluorowodorowego można obliczyć jako 3,46. Gdy stężenie słabego kwasu i sprzężonej zasady w roztworze jest równe, pH jest równe pKa. Dlatego początkowe pH wynosi 3,46.
Jeśli 0,2 mola kwasu solnego zostanie dodane do jednego litra tego buforu, przy założeniu, że powoduje to znikomą zmianę objętości, dodany kwas jest neutralizowany przez jony fluorkowe, wytwarzając kwas fluorowodorowy. Prowadzi to do stechiometrycznego spadku stężenia jonów fluorkowych o 0,2 mola i równego wzrostu stężenia kwasu fluorowodorowego.
Nowe stężenia można wprowadzić do równania Hendersona-Hasselbalcha. Po rozwiązaniu nowe pH wynosi 3,37, czyli mniej niż początkowa wartość pH wynosząca 3,46.
Po dodaniu mocnego kwasu następuje spadek pH, ale spadek jest niewielki, ponieważ roztwór jest buforowany.
W przeciwieństwie do tego, jeśli 0,1 mola wodorotlenku sodu zostanie dodane do jednego litra buforu, przy założeniu, że powoduje to znikomą zmianę objętości, dodana zasada jest neutralizowana przez reakcję z kwasem fluorowodorowym. Powoduje to stechiometryczny spadek stężenia kwasu fluorowodorowego o 0,1 mola i równy wzrost jonów fluorkowych.
Korzystając z równania Hendersona-Hasselbalcha, pH roztworu wynosi 3,50, czyli jest nieco wyższe niż początkowa wartość pH 3,46 ze względu na dodanie wodorotlenku sodu.

obliczanie zmian ph w roztworze buforowym

Bufor może zapobiec nagłemu spadkowi lub wzrostowi pH roztworu po dodaniu mocnego kwasu lub zasady, aż do osiągnięcia jego zdolności buforującej. Ten dodatek mocnego kwasu lub zasady powoduje jednak niewielką zmianę pH roztworu. Taką zmianę pH można obliczyć wyznaczając wynikającą z niej zmianę stężenia składników buforu, czyli słabego kwasu i jego sprzężonej zasady lub odwrotnie. Stężenia uzyskane na podstawie obliczeń stechiometrycznych można wykorzystać do określenia końcowego pH roztworu za pomocą równania Hendersona-Hasselbalcha lub tabeli ICE.

Na przykład buforowany roztwór zawiera 0,65 mola kwasu mrówkowego i mrówczanu sodu. Ponieważ stężenie słabego kwasu i jego sprzężonej zasady jest tutaj takie samo, pH roztworu jest równe pKa słabego kwasu, które w tym przypadku wynosi 3,74. Jeśli do tego roztworu doda się 0,05 mola HNO3, powstałe zmiany w stężeniu kwasu mrówkowego i mrówczanu sodu można określić za pomocą obliczeń stechiometrycznych, jak pokazano w poniższej tabeli.

<div class="overflow">
<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;">H+ (aq)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">HCOO− (aq)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">HCOOH (aq)  </td>
		</tr>
		<tr>
			<td style="color: black; border: 1px solid black; text-align: center;">Przed dodaniem (M)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">~0.00 mol</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0.65 mol</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0.65 mol</td>
		</tr>
		<tr>
			<td style="color: black; border: 1px solid black; text-align: center;">Dodatek (M)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0.050 mol</td>
			<td style="color: black; border: 1px solid black; text-align: center;">-</td>
			<td style="color: black; border: 1px solid black; text-align: center;">-</td>
		</tr>
		<tr>
			<td style="color: black; border: 1px solid black; text-align: center;">Po dodaniu (M)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">~0.00 mol</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0.60 mol</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0.70 mol</td>
		</tr>
	</tbody>
</table>
</div>

Następnie można określić końcowe pH roztworu, wstawiając zmienione stężenia kwasu mrówkowego i mrówczanu sodu do równania Hendersona-Hasselbalcha.

<img alt="Eq1" src="/files/ftp_upload/11411/11411_Equation_1.png" style="text-align: center; height: 165px;" />

Zatem dodatek 0,05 mola HNO3 obniża pH roztworu z 3,74 do 3,67.

Podobnie, jeśli do tego samego roztworu dodane zostanie 0,10 mola NaOH, powstałe zmiany w stężeniu kwasu mrówkowego i mrówczanu sodu można określić za pomocą obliczeń stechiometrycznych, jak pokazano w poniższej tabeli.

<div class="overflow">
<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;">OH− (aq) </td>
			<td style="color: black; border: 1px solid black; text-align: center;">HCOOH (aq)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">HCOO− (aq)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">H2O (l)</td>
		</tr>
		<tr>
			<td style="color: black; border: 1px solid black; text-align: center;">Przed dodaniem (M)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">~0.00 mol </td>
			<td style="color: black; border: 1px solid black; text-align: center;">0.65 mol</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0.65 mol</td>
			<td style="color: black; border: 1px solid black; text-align: center;">-­</td>
		</tr>
		<tr>
			<td style="color: black; border: 1px solid black; text-align: center;">Dodatek (M)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0.10 mol</td>
			<td style="color: black; border: 1px solid black; text-align: center;">-</td>
			<td style="color: black; border: 1px solid black; text-align: center;">-</td>
			<td style="color: black; border: 1px solid black; text-align: center;">­-</td>
		</tr>
		<tr>
			<td style="color: black; border: 1px solid black; text-align: center;">Po dodaniu (M)</td>
			<td style="color: black; border: 1px solid black; text-align: center;">~0.00 mol</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0.55 mol</td>
			<td style="color: black; border: 1px solid black; text-align: center;">0.75 mol</td>
			<td style="color: black; border: 1px solid black; text-align: center;">­-</td>
		</tr>
	</tbody>
</table>
</div>

Następnie można określić końcowe pH roztworu, wstawiając zmienione stężenia kwasu mrówkowego i mrówczanu sodu do równania Hendersona-Hasselbalcha.

<img alt="Eq2" src="/files/ftp_upload/11411/11411_Equation_2.png" style="text-align: center; height: 165px;" />

Zatem dodatek 0,10 mola NaOH zwiększa pH roztworu z 3,74 do 3,87.

Obliczanie zmian pH w roztworze buforowym

Bufor może zapobiec nagłemu spadkowi lub wzrostowi pH roztworu po dodaniu mocnego kwasu lub zasady do jego zdolności buforowej; jednak...

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buffer ph after acid or base addition

A buffer solution can resist sudden pH changes when a strong acid or strong base is added. It does not stop the pH from changing completely. The shift is usually small, as long as the addition stays within the buffer’s capacity.

To find the new pH, first determine how the added acid or base changes the amounts of the buffer components. A buffer contains a weak acid and its conjugate base, or the reverse pair. These amounts are found with stoichiometric calculations, which track how much of each substance reacts.

The updated concentrations can then be used in the Henderson-Hasselbalch equation. An ICE table can also be used. In the example, the buffer contains 0.65 mol of formic acid and 0.65 mol of sodium formate. Because the weak acid and conjugate base are equal, the pH starts at the pKa of formic acid, which is 3.74.

When 0.05 mol HNO3 is added, the hydrogen ions react with formate. After the reaction, the solution contains 0.60 mol sodium formate and 0.70 mol formic acid. Using these new values in the Henderson-Hasselbalch equation gives a final pH of 3.67, down from 3.74.

The same method works when a strong base is added. If 0.10 mol NaOH is added to the same buffer, hydroxide reacts with formic acid. The amounts change to 0.55 mol formic acid and 0.75 mol sodium formate. The final pH then rises to 3.87.

Buffer pH After Acid or Base Addition

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Buffer solutions are mixtures that help keep pH steady. A buffer contains an appreciable amount of a weak conjugate acid-base pair. It resists a change in pH when small amounts of a strong acid or a strong base are added. One common buffer uses a weak acid and its salt. Acetic acid and sodium acetate form this type of system, written as CH3COOH(aq) + CH3COONa(aq). Another buffer uses a weak base and its salt. A solution of ammonia and ammonium chloride, NH3(aq) + NH4Cl(aq), is an example. The...

calculating ph changes in a buffer solution

A buffer solution can resist sudden pH changes when a strong acid or strong base is added. It does not stop the pH from changing completely. The shift is usually small, as long as the addition stays within the buffer’s capacity. To find the new pH, first determine how the added acid or base changes the amounts of the buffer components. A buffer contains a weak acid and its conjugate base, or the reverse pair. These amounts are found with stoichiometric calculations, which track how much of...

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pH at key points in a weak acid titration can be calculated for an acetic acid solution as sodium hydroxide is added. The example uses 25.00 mL of 0.100 M CH3CO2H titrated with 0.100 M NaOH. The reaction is shown as the weak acid reacts with the strong base. The initial pH is found from the acetic acid solution using the usual ICE approach. ICE means initial, change, and equilibrium, and it is a common way to set up equilibrium problems. The solution pH is calculated before any titrant is...

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The Henderson-Hasselbalch equation helps relate buffer pH to the weak acid in the solution. It comes from the ionization-constant expression for a weak acid. By rearranging that expression to solve for hydronium ion concentration, [H3O+], and then taking the negative logarithm of both sides, the equation can be written in terms of pH. In this form, pKa is defined as the negative logarithm of the weak acid ionization constant, or pKa = −log Ka. The equation shows the connection between pH, pKa,...

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Acid-base titration uses pH changes to track how a solution responds as titrant is added. A titration curve is a graph of a solution property versus the amount of added titrant. In acid-base titrations, pH is the main property to follow because it changes in a predictable way with the solution makeup and helps show the endpoint. This pattern can be seen with a strong acid and strong base, a strong acid and weak base, or a strong base and weak acid. For example, 25.00 mL of 0.100 M HCl titrated...

key terms and concepts

Learn how buffers resist pH change after strong acid or base is added. Use stoichiometry and the Henderson-Hasselbalch equation to find the new pH.

Buffer pH After Acid or Base Addition | JoVE K12

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• Energy
• Equilibrium
• Acids/bases

Content elaboration: In the Physical Science course, coefficients were used to balance simple equations. Other representations, including Lewis structures and three-dimensional models, were also used and manipulated to demonstrate the conservation of matter in chemical reactions. In this course, more complex reactions will be studied, classified and represented with balanced chemical equations and three-dimensional models. Classifying reactions into types can be a helpful organizational tool for recognizing patterns of what may happen when two substances are mixed. Teachers should be aware that the common reaction classifications that are often used in high school chemistry courses may lead to misconceptions because they are not based on the actual chemistry, but on surface features that can be similar from one system to another (e.g., exchanging partners), even though the underlying chemistry is not the same. However, these classifications may be useful in making predictions about what happens when two substances are mixed. Some general types of chemical reactions are oxidation/reduction, synthesis, decomposition, single replacement, double replacement (including precipitation reactions and some acid-base neutralizations) and combustion reactions. Some reactions can fit into more than one category. For example, a single replacement reaction can also be classified as an oxidation/reduction reaction. Identification of reactions involving oxidation and reduction as well as indicating what substance is being oxidized and what is being reduced are appropriate in this course. However, balancing complex oxidation/reduction reactions is reserved for more advanced study. Organic molecules release energy when undergoing combustion reactions and are used to meet the energy needs of society (e.g., oil, gasoline, natural gas) and to provide the energy needs of biological organisms (e.g., cellular respiration). When a reaction between two ionic compounds in aqueous solution results in the formation of a precipitate or molecular compound, the reaction often occurs because the new ionic or covalent bonds are stronger than the original ion-dipole interactions of the ions in solution. Laboratory experiences (3-D or virtual) with different types of chemical reactions should be provided. Reactions occur when reacting particles collide in an appropriate orientation and with sufficient energy. The rate of a chemical reaction is the change in the amount of the reactants or products in a specific period of time. Increasing the probability or effectiveness of the collisions between the particles increases the rate of the reaction. Therefore, changing the concentration of the reactants, changing the temperature or the pressure of gaseous reactants, or using a catalyst, can change the reaction rate. Likewise, the collision theory can be applied to dissolving solids in a liquid solvent and can be used to explain why reactions are more likely to occur between reactants in the aqueous or gaseous state than between solids. The rate at which a substance dissolves should not be confused with the amount of solute that can dissolve in a given amount of solvent (solubility). Mathematical treatment of reaction rates is reserved for more advanced study. Computer simulations can help visualize reactions from the perspective of the kinetic-molecular theory. In middle school, the differences between potential and kinetic energy and the particle nature of thermal energy were introduced. For chemical systems, potential energy is in the form of chemical energy and kinetic energy is in the form of thermal energy. The total amount of chemical energy and/or thermal energy in a system is impossible to measure. However, the energy change of a system can be calculated from measurements (mass and change in temperature) from calorimetry experiments in the laboratory. Conservation of energy is an important component of calorimetry equations. Thermal energy is the energy of a system due to the movement of its particles. The thermal energy of an object depends upon the temperature and chemical composition. Some materials require little energy to change their temperature and other materials require a great deal to change their temperature by the same amount. Specific heat is a measure of how much energy is needed to change the temperature of a specific mass of material a specific amount. Specific heat values can be used to calculate the thermal energy change, the temperature (initial, final or change in) or mass of a material in calorimetry. Water has a particularly high specific heat capacity, which is important in regulating Earth's temperature. As studied in middle school, chemical energy is the potential energy associated with chemical systems. Chemical reactions involve valence electrons forming bonds to yield more stable products with lower energies. Energy is required to break interactions and bonds between the reactant atoms and energy is released when an interaction or bond is formed between the atoms in the products. Molecules with weak bonds (e.g., ATP) are less stable and tend to react to produce more stable products, releasing energy in the process. Generally, energy is transferred out of the system (exothermic) when the products have stronger bonds than the reactants and is transferred into the system (endothermic) when the reactants have stronger bonds than the products. Predictions of the energy requirements (endothermic or exothermic) of a reaction can be made given a table of bond energies. Graphic representations can be drawn and interpreted to represent the energy changes during a reaction. The role of energy in determining the spontaneity of chemical reactions is dealt with conceptually in this course. Entropy and its influence on the spontaneity of reactions are reserved for more advanced study. All reactions are reversible to a degree and many reactions do not proceed completely toward products but appear to stop progressing before the reactants are all used up. At this point, the amounts of the reactants and the products appear to be constant and the reaction can be said to have reached dynamic equilibrium. Dynamic equilibrium means the rate of the reverse reaction is equal to the rate of the forward reaction so there is no apparent change in the reaction. If a chemical system at equilibrium is disturbed by a change in the conditions of the system (e.g., increase or decrease in the temperature, pressure on gaseous equilibrium systems, concentration of a reactant or product), then the equilibrium system will respond by shifting to a new equilibrium state, reducing the effect of the change (Le Chatelier's Principle). If products are removed as they are formed during a reaction, then the equilibrium position of the system is forced to shift to favor the products. In this way, an otherwise unfavorable reaction can be made to occur. Mathematical treatment of equilibrium reactions is reserved for advanced study. Computer simulations can help visualize the progression of a reaction to dynamic equilibrium and the continuation of both the forward and reverse reactions after equilibrium has been attained. Properties of acids and bases and the ranges of the pH scale were introduced in Physical Science. In this course, the structural features of molecules are explored to further understand acids and bases. Acids often result when hydrogen is covalently bonded to an electronegative element and is easily dissociated from the rest of the molecule to bind with water to form a hydronium ion (H3O+). The acidity of an aqueous solution can be expressed as pH, where pH can be calculated from the concentration of the hydronium ion. Bases are likely to dissociate in water to form a hydroxide ion. Acids can react with bases to form a salt and water. Such neutralization reactions can be studied quantitatively by performing titration experiments.

Buffer pH After Acid or Base Addition

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