Ka and Kb calculations
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Topic Notes
Introduction to Ka and Kb Calculations
Ka and Kb calculations are fundamental concepts in chemistry, dealing with acid dissociation constants for weak acids and bases. The introduction video provides a comprehensive overview of these calculations, serving as a crucial starting point for understanding the topic. Ka represents the acid dissociation constant, while Kb is the base dissociation constant. These values are essential in determining the strength of weak acids and bases, respectively. By mastering Ka and Kb calculations, students can predict the behavior of acids and bases in solution, calculate pH levels, and understand buffer systems. These calculations play a vital role in various chemical processes, from industrial applications to biological systems. The ability to perform and interpret Ka and Kb calculations is indispensable for chemists, allowing them to analyze and manipulate acid-base equilibria in diverse scientific and practical contexts.
Understanding Ka Calculations for Weak Acids
The acid dissociation constant, commonly known as Ka, is a fundamental concept in chemistry that plays a crucial role in understanding the behavior of weak acids in solution. Ka is a quantitative measure of the strength of an acid in solution, representing the extent to which an acid dissociates into its constituent ions. This constant is particularly important when dealing with weak acids, as it allows chemists to predict and calculate various properties of acidic solutions, including pH.
The Ka expression is derived from the equilibrium reaction of a weak acid (HA) dissociating into its conjugate base (A-) and a hydrogen ion (H+) in an aqueous solution. The general equation for this dissociation is:
HA + H2O H3O+ + A-
The Ka equation is then expressed as:
Ka = [H+][A-] / [HA]
Where [H+] is the concentration of hydrogen ions, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the undissociated acid. The square brackets denote molar concentrations.
Understanding the relationship between Ka and pH is crucial for chemists and students alike. The pH of a solution is directly related to the concentration of hydrogen ions, and for weak acids, this concentration can be calculated using the Ka value. Here's a step-by-step guide on how to calculate pH from Ka, using methanoic acid (HCOOH) as an example:
Step 1: Write the Ka expression for the acid.
For methanoic acid: Ka = [H+][HCOO-] / [HCOOH]
Step 2: Substitute the known values. Let's assume we have a 0.1 M solution of methanoic acid with a Ka of 1.8 × 10^-4.
Step 3: Set up an ICE table (Initial, Change, Equilibrium) to track the concentrations of each species.
Step 4: Use the Ka expression to set up an equation.
1.8 × 10^-4 = x^2 / (0.1 - x), where x represents [H+] at equilibrium.
Step 5: Solve for x using the quadratic equation or by making assumptions (discussed later).
Step 6: Once you have [H+], calculate the pH using the equation: pH = -log[H+]
In Ka calculations, certain assumptions are often made to simplify the process. The most common assumption is that for weak acids, the amount of acid that dissociates is negligible compared to the initial concentration. This allows us to approximate (0.1 - x) as simply 0.1 in our example. This assumption is generally valid when the Ka value is less than 10^-3 and the acid concentration is greater than 0.01 M.
The justification for this assumption lies in the fact that weak acids dissociate only to a small extent. For instance, if less than 5% of the acid dissociates, the error introduced by this assumption is typically less than 5%, which is often acceptable for many calculations.
Another important aspect of Ka calculations is the consideration of the autoionization of water, especially for very weak acids or dilute solutions. In such cases, the contribution of H+ ions from water itself may become significant and should be accounted for in more precise calculations.
Understanding Ka and its relationship to pH is essential for various applications in chemistry, including buffer preparation, acid-base titrations, and predicting the behavior of weak acids in different environments. It allows chemists to design experiments, formulate products, and understand biological processes where acid-base equilibria play a crucial role.
In conclusion, the acid dissociation constant (Ka) is a powerful tool in chemistry for understanding and quantifying the behavior of weak acids. By mastering Ka calculations and understanding the assumptions involved, chemists can accurately predict and control the pH of solutions, which is crucial in numerous scientific and industrial applications. Whether you're
Kb Calculations for Weak Bases
In chemistry, understanding acid-base equilibria is crucial, and two important constants play a significant role: Ka (acid dissociation constant) and Kb (base dissociation constant). While Ka is more commonly discussed, Kb is equally important, especially when dealing with weak bases. This article will explore the concept of Kb, its relationship to Ka, and how to perform Kb calculations for weak bases.
Kb, or the base dissociation constant, is a measure of how completely a base dissociates in an aqueous solution. It is the equilibrium constant for base dissociation. The Kb equation represents this equilibrium:
B + H2O BH+ + OH-
Where B is the base, BH+ is the conjugate acid, and OH- is the hydroxide ion. The Kb equation is expressed as:
Kb = [BH+][OH-] / [B]
Understanding how to find Kb is essential for solving problems involving weak bases. Interestingly, Kb can be determined indirectly using Ka and Kw (the ion product constant of water). This relationship is expressed as:
Ka × Kb = Kw
Where Kw is approximately 1.0 × 10^-14 at 25°C. This equation highlights the relationship between Ka and Kb. When comparing Kb vs Ka, it's important to note that a strong base will have a large Kb value, while its conjugate acid will have a small Ka value, and vice versa.
To illustrate a Kb calculation, let's consider a weak base example using ammonia (NH3). The Kb of ammonia is 1.8 × 10^-5 at 25°C. We'll calculate the pH of a 0.1 M ammonia solution.
Step 1: Write the equilibrium equation
NH3 + H2O NH4+ + OH-
Step 2: Set up the ICE table (Initial, Change, Equilibrium)
Initial: [NH3] = 0.1 M, [NH4+] = 0 M, [OH-] = 0 M
Change: -x, +x, +x
Equilibrium: (0.1 - x), x, x
Step 3: Apply the Kb equation
Kb = [NH4+][OH-] / [NH3] = 1.8 × 10^-5
(x)(x) / (0.1 - x) = 1.8 × 10^-5
Step 4: Assume x is small compared to 0.1, so (0.1 - x) 0.1
x^2 / 0.1 = 1.8 × 10^-5
x^2 = 1.8 × 10^-6
x = (1.8 × 10^-6) = 1.34 × 10^-3
Step 5: Calculate [OH-]
[OH-] = x = 1.34 × 10^-3 M
Step 6: Calculate pOH
pOH = -log[OH-] = -log(1.34 × 10^-3) = 2.87
Step 7: Calculate pH
pH + pOH = 14
pH = 14 - pOH = 14 - 2.87 = 11.13
This example demonstrates the process of a Kb calculation for a weak base. It's important to note the differences between Ka and Kb calculations. While the general approach is similar, there are pOH calculation for weak bases.
Relationship Between Ka, Kb, and Kw
Understanding the relationship between Ka, Kb, and Kw is crucial in acid-base chemistry. These constants play a vital role in determining the strength of acids and bases, as well as in various chemical calculations. Let's explore how these constants are interconnected and how to use them effectively.
Ka, or the acid dissociation constant, represents the strength of an acid in water. Similarly, Kb, the base dissociation constant, indicates the strength of a base. Kw, known as the ionization constant of water, is a fundamental constant that represents the product of hydrogen ion and hydroxide ion concentrations in water at a given temperature.
The relationship between these constants is expressed by the equation: Ka * Kb = Kw. This formula is essential for converting between Ka and Kb values. To find Ka from Kb or vice versa, you can use the Ka and Kb equation: Ka = Kw / Kb or Kb = Kw / Ka. At 25°C, Kw has a value of 1.0 x 10^-14, which simplifies these calculations.
For example, if you need to find Ka from Kb, and you know that Kb = 1.0 x 10^-5, you can calculate Ka as follows:
Ka = Kw / Kb = (1.0 x 10^-14) / (1.0 x 10^-5) = 1.0 x 10^-9
This Ka and Kb formula is particularly useful when dealing with conjugate acid-base pairs. A conjugate acid-base pair consists of two species that differ by a single proton. The acid in one pair is the conjugate acid of the base in the other pair, and vice versa. Understanding this relationship is crucial for solving problems involving weak acids and bases.
When working with conjugate pairs, you can use the Ka Kb equation to find unknown values. For instance, if you know the Ka of a weak acid, you can calculate the Kb of its conjugate base using the relationship Kb = Kw / Ka. This is particularly helpful in problems where you need to determine the pH or pOH of a solution.
The concept of Kb to pOH conversion is another important application of these relationships. pOH is the negative logarithm of the hydroxide ion concentration, and it's related to Kb through the equation: pOH = -log[OH-] = -log((Kb * Cb)), where Cb is the initial concentration of the base.
In practical terms, these relationships are used in various chemical calculations, such as buffer preparation, titration analysis, and predicting the behavior of acids and bases in solution. For example, when designing a buffer solution, knowing the Ka of the weak acid component allows you to calculate the Kb of its conjugate base, which is essential for determining the buffer's effectiveness.
Moreover, understanding how to find Ka from Kb (and vice versa) is crucial in pharmaceutical research, environmental science, and industrial processes. It allows scientists and engineers to predict the behavior of compounds in different pH environments, which is vital for drug development, water treatment, and manufacturing processes.
In conclusion, the relationship between Ka, Kb, and Kw is a fundamental concept in acid-base chemistry. The ability to convert between these constants and understand their implications is essential for solving a wide range of chemical problems. By mastering these relationships and the associated calculations, you'll be well-equipped to tackle complex acid-base equilibria and their applications in various scientific and industrial fields.
Practical Applications and Problem-Solving
Understanding Ka and Kb calculations is crucial for solving various acid-base problems in chemistry. Let's explore practical examples and problem-solving techniques for these calculations, focusing on finding pH from Ka and Kb, calculating percentage dissociation, and determining acid and base strength.
How to Determine pH from Ka
To find the pH of a weak acid solution using Ka, follow these steps:
- Write the acid dissociation equation
- Set up the Ka expression
- Use the ICE table (Initial, Change, Equilibrium)
- Solve for the equilibrium concentration of H+
- Calculate pH using pH = -log[H+]
Example: Calculate the pH of a 0.1 M acetic acid solution (Ka = 1.8 × 10^-5)
Step 1: CH3COOH H+ + CH3COO-
Step 2: Ka = [H+][CH3COO-] / [CH3COOH]
Step 3: ICE table
Step 4: 1.8 × 10^-5 = x^2 / (0.1 - x), assume x << 0.1
x = (1.8 × 10^-5 × 0.1) = 1.34 × 10^-3 M
Step 5: pH = -log(1.34 × 10^-3) = 2.87
Finding pH from Kb
The process for calculating pH from Kb is similar, but we focus on pOH first:
- Write the base dissociation equation
- Set up the Kb expression
- Use the ICE table
- Solve for the equilibrium concentration of OH-
- Calculate pOH using pOH = -log[OH-]
- Find pH using pH + pOH = 14
Example: Calculate the pH of a 0.05 M ammonia solution (Kb = 1.8 × 10^-5)
Step 1: NH3 + H2O NH4+ + OH-
Step 2: Kb = [NH4+][OH-] / [NH3]
Step 3: ICE table
Step 4: 1.8 × 10^-5 = x^2 / (0.05 - x), assume x << 0.05
x = (1.8 × 10^-5 × 0.05) = 9.49 × 10^-4 M
Step 5: pOH = -log(9.49 × 10^-4) = 3.02
Step 6: pH = 14 - 3.02 = 10.98
Calculating Percentage Dissociation
Percentage dissociation is a measure of how much an acid or base dissociates in solution. To calculate it:
- Find the equilibrium concentration of the dissociated species
- Divide by the initial concentration of the acid or base
- Multiply by 100
Example: Calculate the percent dissociation of acetic acid in the previous example
Percent dissociation = (1.34 × 10^-3 M / 0.1 M) × 100 =
Common Mistakes and Troubleshooting
When performing Ka and Kb calculations, students often encounter challenges that can lead to significant errors in pH calculations. Understanding these common mistakes and learning how to troubleshoot them is crucial for mastering acid-base chemistry. One frequent error is misinterpreting the Ka to pH formula or struggling with how to go from Ka to pH. Students may forget that Ka represents the acid dissociation constant, while Kb is for bases. This confusion can result in incorrect pH values, especially when dealing with weak acids or bases.
Another common pitfall is neglecting to consider the initial concentrations of acids or bases. When using the Ka and pH relationship, it's essential to account for the starting concentration of the acid. Similarly, for Kb to pOH calculations, the initial base concentration plays a vital role. Overlooking these factors can lead to substantial errors in the final pH or pOH values.
Students often struggle with unit conversions, particularly when working with logarithmic scales. Remember that pH is the negative logarithm of hydrogen ion concentration, while pOH is the negative logarithm of hydroxide ion concentration. Failing to convert between these units correctly can result in pH calculation errors that are off by orders of magnitude.
To avoid these mistakes, it's crucial to develop a strong understanding of the underlying concepts rather than simply memorizing formulas. Start by clearly identifying whether you're dealing with an acid (Ka) or a base (Kb). Then, carefully set up your ICE table (Initial, Change, Equilibrium) to track concentration changes. Double-check your units and ensure you're using the correct logarithmic relationships.
When troubleshooting, always verify your work step-by-step. A small error in the initial setup can propagate through the entire calculation. For instance, misplacing a decimal point when entering Ka values can lead to pH results that are completely off. Similarly, forgetting to square a concentration in equilibrium expressions can significantly alter the outcome.
To illustrate, consider a weak acid with Ka = 1.8 × 10-5. If a student mistakenly uses 1.8 × 10-4, the calculated pH could be off by nearly an entire unit. This seemingly small error in transcribing the Ka value leads to a tenfold difference in the hydrogen ion concentration, dramatically affecting the solution's acidity.
Practice is key to mastering these calculations. Work through a variety of problems, from simple Ka to pH conversions to more complex buffer systems. Always check your answers for reasonableness remember that most aqueous solutions have pH values between 0 and 14. By focusing on understanding the principles behind acid-base equilibria and carefully applying the correct formulas, you'll improve your accuracy and confidence in pH calculations.
Conclusion
In this article, we've explored the crucial concepts of Ka and Kb calculations in acid-base chemistry. Understanding these principles is fundamental for mastering pH calculations and predicting chemical reactions. We've covered the key formulas, step-by-step problem-solving techniques, and practical applications of Ka and Kb in various chemical scenarios. To solidify your grasp of these concepts, we encourage you to rewatch the introductory video, which provides a comprehensive overview of the topic. Remember, proficiency in Ka and Kb calculations is essential for success in advanced chemistry courses and real-world applications. To further enhance your skills, practice solving a variety of problems and engage with additional resources. Don't hesitate to ask questions and seek clarification on challenging aspects. By mastering these calculations, you'll gain a deeper understanding of acid-base chemistry and be well-prepared for more complex chemical analyses. Keep exploring and expanding your knowledge in this fascinating field!
Example:
Applying the Ka expression Weak acids/bases at equilibrium.
Step 1: Introduction to Ka and Kb Calculations
Welcome to the section on Ka and Kb calculations. In this guide, we will walk through the process of finding the pH of a solution given the acid dissociation constant (Ka) and the concentration of a weak acid. We will also touch upon the assumptions made during these calculations and how to solve for Kb using the Kw expression.
Step 2: Understanding Weak Acids and Bases at Equilibrium
Weak acids and bases do not fully dissociate in water. This partial dissociation means that the equilibrium expression for weak acids must account for both the undissociated acid and the ions produced. For weak acids, the general dissociation can be represented as:
HX + H2O H3O+ + X-
Here, HX is the weak acid, H3O+ is the hydronium ion, and X- is the conjugate base. The Ka expression for this equilibrium is:
Ka = [H3O+][X-] / [HX]
Step 3: Setting Up the Equilibrium Expression
To find the pH of a weak acid solution, we start by setting up the equilibrium expression. Consider a weak acid HX with an initial concentration of 'a'. At equilibrium, the concentration of HX will be 'a - b', where 'b' is the amount that dissociates. The concentrations of H3O+ and X- at equilibrium will both be 'b'. Thus, the Ka expression becomes:
Ka = [H3O+][X-] / [HX] = b2 / (a - b)
Step 4: Making Assumptions for Simplification
Given that weak acids do not dissociate completely, 'b' is typically much smaller than 'a'. This allows us to make the assumption that 'a - b' is approximately equal to 'a'. This simplifies the Ka expression to:
Ka b2 / a
This assumption holds true for most weak acids, where the degree of dissociation is less than 50%.
Step 5: Solving for the Hydronium Ion Concentration
To find the concentration of H3O+ (b), we rearrange the simplified Ka expression:
b = (Ka * a)
By substituting the known values of Ka and 'a', we can calculate the concentration of H3O+.
Step 6: Calculating the pH
Once we have the concentration of H3O+, we can find the pH of the solution using the formula:
pH = -log[H3O+]
Substitute the value of [H3O+] obtained from the previous step to calculate the pH.
Step 7: Example Calculation
Let's consider an example with 0.1 M methanoic acid (HCOOH) with a Ka of 1.8 x 10-4. Using the simplified Ka expression:
b = (1.8 x 10-4 * 0.1) = (1.8 x 10-5) 4.24 x 10-3
Then, the pH is calculated as:
pH = -log(4.24 x 10-3) 2.37
Step 8: Conclusion
By following these steps, you can apply the Ka expression to find the pH of a weak acid solution. Remember to consider the assumptions made during the calculations and verify if they hold true for your specific case. This method can also be adapted for weak bases using the Kb expression and the relationship between Ka, Kb, and Kw.
FAQs
Here are some frequently asked questions about Ka and Kb calculations:
1. How do you calculate pH from Ka?
To calculate pH from Ka, follow these steps:
- Write the acid dissociation equation
- Set up the Ka expression
- Use an ICE table to solve for [H+]
- Calculate pH using the formula: pH = -log[H+]
For example, if Ka = 1.8 × 10-5 for a 0.1 M weak acid solution:
Ka = [H+][A-] / [HA] = 1.8 × 10-5 = x2 / (0.1 - x)
Solve for x (assuming x << 0.1), then calculate pH = -log(x)
2. What is the relationship between Ka and Kb?
The relationship between Ka and Kb is expressed by the equation:
Ka × Kb = Kw
Where Kw is the ion product constant of water (1.0 × 10-14 at 25°C). This relationship allows you to calculate Kb if you know Ka, and vice versa.
3. How do you convert from Ka to Kb?
To convert from Ka to Kb, use the following formula:
Kb = Kw / Ka
For example, if Ka = 1.8 × 10-5, then:
Kb = (1.0 × 10-14) / (1.8 × 10-5) = 5.56 × 10-10
4. How is pKa related to Ka?
pKa is the negative logarithm of Ka:
pKa = -log(Ka)
This relationship is useful for comparing acid strengths. A lower pKa indicates a stronger acid.
5. How do you calculate the percentage dissociation of a weak acid?
To calculate the percentage dissociation:
- Find the equilibrium concentration of H+ ions
- Divide by the initial concentration of the acid
- Multiply by 100
Percentage dissociation = ([H+] / [HA]initial) × 100
For example, if [H+] = 1.34 × 10-3 M for a 0.1 M acid solution:
Percentage dissociation = (1.34 × 10-3 / 0.1) × 100 = 1.34%
Prerequisite Topics for Ka and Kb Calculations
Understanding Ka and Kb calculations is crucial in chemistry, particularly when dealing with acids and bases. However, to truly grasp these concepts, it's essential to have a solid foundation in certain prerequisite topics. Two key areas that are fundamental to mastering Ka and Kb calculations are the acid dissociation constant and strong and weak acids and bases.
The acid dissociation constant, also known as Ka, is a fundamental concept that quantifies the strength of an acid in solution. This constant is directly related to the equilibrium constant for base dissociation, which is essential for understanding Kb calculations. By mastering the acid dissociation constant, students can more easily comprehend how acids behave in solution and how this relates to their strength and dissociation properties.
Similarly, a thorough understanding of strong and weak acids and bases is crucial for Ka and Kb calculations. This topic provides the foundation for distinguishing between different types of acids and bases based on their dissociation behavior. It's particularly important when dealing with weak bases, as pOH calculations for weak bases are directly related to Kb values.
The relationship between these prerequisite topics and Ka and Kb calculations is intricate and multifaceted. For instance, the acid dissociation constant (Ka) is used to determine the strength of an acid, which in turn affects its behavior in solution. This knowledge is crucial when performing calculations involving Ka values. Similarly, understanding the characteristics of strong and weak acids and bases helps in predicting their behavior and applying the appropriate formulas in Ka and Kb calculations.
Moreover, these prerequisite topics provide the conceptual framework necessary for interpreting Ka and Kb values. For example, knowing how to classify acids and bases as strong or weak based on their dissociation constants allows students to make informed predictions about their behavior in various chemical reactions and equilibrium situations.
In conclusion, a solid grasp of the acid dissociation constant and strong and weak acids and bases is essential for success in Ka and Kb calculations. These prerequisite topics provide the necessary foundation for understanding the underlying principles, applying the correct formulas, and interpreting results accurately. By mastering these concepts, students will be well-equipped to tackle more complex problems involving acid-base equilibria and related calculations in their chemistry studies.
In this lesson, we will learn:
- How to apply the Ka expression to find the pH of weak acid solutions.
- The assumptions made in Ka calculations at equilibrium and how to justify them.
- How to solve for Kb using Kw and find the pH of weak base solutions.
Notes:
- We know that weak acids and bases are any acid/base species that does not completely dissociate in water. Dissolving a weak acid in water then has two chemical effects:
- Some of the weak acid, HX, will interact with water and dissociate into H3O+ and X- ions.
- The rest of the HX will stay un-dissociated.
HX + H2O ⇌ H3O+ + Cl-
Start concentration (M)
Equilibrium conc. (M)
HX
A
A - B
H3O+
≈ 0
B
X-
0
B
The acid dissociation constant, Ka can be expressed in these terms:Ka = [HX][X−][H3O+] = [A]−B[X2] - Taking an example with 0.1M methanoic acid HCOOH, we can write the following:
Start concentration (M)
Equilibrium conc. (M)
HCOOH
0.1
0.1 - B
H3O+
0
B
X-
0
B
We can apply the acid dissociation constant, Ka to this equilibrium. Methanoic acid1 has a Ka value of 1.8∗10-4 so the Ka equation can be written fully:Ka = [HCOOH][HCOO−][H3O+] = [0.1]−[H3O+][H3O+]2 = 1.8 ∗ 10-4
Some assumptions are made to complete this calculation:- The starting concentration of H3O+ ions, in neutral water is only 1*10-7 M (see Autoionization of water) and an equally tiny amount of hydroxide ions are also present. This is an incredibly small amount, so this H3O+ is not taken into the calculation; only H3O+ due to the weak acid is used in the calculation.
- With weak acids, we assume that the acid is weak enough that the amount of dissociation doesn’t affect acid concentration. Using the table above, 0.1 M HCOOH added to neutral water will still have concentration of approximately 0.1M at equilibrium, and we can ignore the ‘– B’ in ‘A-B’. Where [HX]eq = equilibrium concentration, [HX]i = start concentration of HX:
[HX]eq - B ≅ [HX]i
YOU MUST STATE THIS ASSUMPTION IN CALCULATIONS.
With the assumptions, we have a final expression:Ka = [HCOOH][HCOO−][H3O+] = [0.1][H3O+]2 = 1.8 ∗ 10-4 0.1 ∗ 1.8 ∗ 10-4 = [H3O+]2 1.8∗10−5 = [H3O+] = 4.24 ∗ 10-3 pH = -log [H3O+] = 2.37
Note that the assumption we made was that [HX] – B is approximately equal to [HX]. We now know that B = 4.24*10-3, so our assumption in this example was to say that 0.1 – 4.24*10-3 = 0.0958.
The assumption can be justified if percentage dissociation is less than 5% which we can work out:
% dissociation = [HX]i[H3O+]eq ∗ 100
Therefore:
% dissociation = [0.1][4.24∗10−3] ∗ 100 = 4.24%
As the calculation shows, the assumption was justified as only 4.24% dissociation occurs. - Kb calculations are similar to Ka calculations with some changes:
- Because acidity strength tables give only Ka, Kb of a weak base will need to be found by the calculation in the autoionization of water expression. You will need to find the Ka of the conjugate acid in the acidity strength table to do this.
- The equilibrium concentrations you obtain using Kb will give you [OH-], so pH will need to be found by solving: pH = 14 – pOH.
Taking an example with 0.5M of the weak base ammonia, NH3, we can write the following:
NH3 + H2O ⇌ NH4+ + OH- Start concentration (M)
Equilibrium conc. (M)
NH3
0.5
0.5 - B
NH3+
0
B
OH-
0
B
The conjugate acid of ammonia is the ammonium ion, NH4+ which has a Ka value1 of 5.6 ∗ 10-10. Solving the autoionization expression for Kb(NH3) gives:Kb = KaKw = 5.6∗10−1010−14 = 1.79 ∗ 10-5
Using our value for Kb (now rounding to 1.8 ∗ 10-5 or 2 significant figures) we can solve for the equilibrium concentration of hydroxide ions. Again, we make the assumption that the concentration of NH3 isn’t significantly affected by the dissociation into NH4+:[NH3]eq - B ≅ [NH3]i
Now we can find the hydroxide ion concentration:Kb = [NH3][NH4+][OH−] = [0.5][OH−]2 = 1.8 ∗ 10-5 [OH-] = ((1.8∗10−5)∗0.5) = 3 ∗ 10-3 pOH = -log[OH-] = -log (3 ∗ 10-3) = 2.52 pH = 14 - pOH = 14 - 2.52 = 11.48
Testing the assumption can now be done:% dissociation = [NH3]i[OH−]eq ∗ 100 = 0.53∗10−3 ∗ 100 = 0.6 %
With only 0.6% dissociation, the assumption is justified and the pH has been found. - Another calculation that shows the difference between strong and weak acids and bases is the effect of dilution on pH.
The effect on strong acids is straightforward, because we assume 100% dissociation:- If a solution of strong acid, e.g. HCl is 1M, then [H3O+ (aq)] = 1M. Taking the negative log of this:
pH = -log[1] = 0
Diluting this by a factor of 10 will give a concentration of 0.1M
pH = -log[0.1] = 1
A further 10, or 100 fold from the original:
pH = -log[0.01] = 2
In short, diluting a strong acid or base has a direct logarithmic effect on pH.
- If a solution of weak acid, e.g. CH3COOH is 1M, then pH and [H3O+ (aq)] is worked out using the Ka expression:
Ka (CH3COOH) = 1.4*10-5Ka = [CH3COOH][H3O+][CH3COO−]
The calculation to find pH using this expression has been explained above, so moving forward to an answer (using the assumptions needed)1.4 * 10-5 = [1][H3O+][CH3COO−] where [H3O+] = [CH3COOH-] 1.4∗10−5 = [H3O+] = 3.74 * 10-3 pH = -log[ 3.74 * 10-3] = 2.42
A dilution of this weak acid solution to make it 0.1M would have the following effect on the calculation:1.4 * 10-5 = [0.1][H3O+][CH3COO−] where [H3O+] = [CH3COOH-] 1.4∗10−6 = [H3O+] = 1.18 * 10-3 pH = -log[ 1.18 * 10-3] = 2.92
Another dilution by a factor of ten:1.4 * 10-5 = [0.01][H3O+][CH3COO−] where [H3O+] = [CH3COOH-] 1.4∗10−7 = [H3O+] = 3.74 * 10-4 pH = -log[ 3.74 * 10-4] = 3.42
In short, diluting a weak acid has a lesser effect on pH than in strong acids.
- If a solution of strong acid, e.g. HCl is 1M, then [H3O+ (aq)] = 1M. Taking the negative log of this:
remaining today
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