Consider The Titration Of 30.0 ML Of 0.267 M Weak Base $B \left( K_b =1.3 \times 10^{-10}\right)$ With 0.150 M HI.What Would Be The PH Of The Solution After The Addition Of 20.0 ML Of HI?

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Introduction

Titration is a laboratory technique used to determine the concentration of a substance by reacting it with a known amount of another substance. In this article, we will explore the titration of a weak base with a strong acid, specifically the titration of 30.0 mL of 0.267 M weak base $B \left( K_b =1.3 \times 10^{-10}\right)$ with 0.150 M HI. We will calculate the pH of the solution after the addition of 20.0 mL of HI.

Weak Bases and Strong Acids

A weak base is a base that does not completely dissociate in water, resulting in a solution with a pH greater than 7. A strong acid, on the other hand, completely dissociates in water, resulting in a solution with a pH less than 7. In this titration, we are dealing with a weak base $B$ and a strong acid HI.

The Titration Process

The titration process involves the addition of a strong acid (HI) to a weak base ($B$) until the base is completely neutralized. The reaction between the weak base and the strong acid can be represented by the following equation:

B+HI→BH++I−B + HI \rightarrow BH^+ + I^-

Calculating the Number of Moles of Weak Base

To calculate the number of moles of weak base $B$, we can use the formula:

n=C×Vn = C \times V

where $n$ is the number of moles, $C$ is the concentration of the weak base (0.267 M), and $V$ is the volume of the weak base (30.0 mL).

n=0.267 M×30.0 mL=0.00801 moln = 0.267 \, M \times 30.0 \, mL = 0.00801 \, mol

Calculating the Number of Moles of HI Added

To calculate the number of moles of HI added, we can use the formula:

n=C×Vn = C \times V

where $n$ is the number of moles, $C$ is the concentration of the strong acid (0.150 M), and $V$ is the volume of the strong acid added (20.0 mL).

n=0.150 M×20.0 mL=0.003 moln = 0.150 \, M \times 20.0 \, mL = 0.003 \, mol

Calculating the pH of the Solution

After the addition of 20.0 mL of HI, the solution contains a mixture of the weak base $B$ and the strong acid HI. To calculate the pH of the solution, we need to consider the equilibrium between the weak base and the strong acid.

The equilibrium constant for the reaction between the weak base and the strong acid is given by the formula:

Kb=[BH+][I−][B]K_b = \frac{[BH^+][I^-]}{[B]}

where $K_b$ is the base dissociation constant (1.3 x 10^-10), $[BH^+]$ is the concentration of the conjugate acid, $[I^-]$ is the concentration of the conjugate base, and $[B]$ is the concentration of the weak base.

Since the strong acid HI is a strong acid, it completely dissociates in water, resulting in a solution with a pH less than 7. Therefore, we can assume that the concentration of the conjugate acid $[BH^+]$ is equal to the concentration of the strong acid HI.

The concentration of the weak base $[B]$ can be calculated using the formula:

[B]=nV[B] = \frac{n}{V}

where $n$ is the number of moles of the weak base (0.00801 mol), and $V$ is the volume of the weak base (30.0 mL + 20.0 mL = 50.0 mL).

[B]=0.00801 mol50.0 mL=0.0001602 M[B] = \frac{0.00801 \, mol}{50.0 \, mL} = 0.0001602 \, M

The concentration of the conjugate acid $[BH^+]$ can be calculated using the formula:

[BH+]=nV[BH^+] = \frac{n}{V}

where $n$ is the number of moles of the strong acid HI (0.003 mol), and $V$ is the volume of the strong acid HI (20.0 mL).

[BH+]=0.003 mol20.0 mL=0.00015 M[BH^+] = \frac{0.003 \, mol}{20.0 \, mL} = 0.00015 \, M

The concentration of the conjugate base $[I^-]$ can be calculated using the formula:

[I−]=[BH+][I^-] = [BH^+]

[I−]=0.00015 M[I^-] = 0.00015 \, M

Now, we can calculate the pH of the solution using the formula:

pH=−log⁡([BH+][I−][B])pH = -\log \left( \frac{[BH^+][I^-]}{[B]} \right)

pH=−log⁡(0.00015 M×0.00015 M0.0001602 M)pH = -\log \left( \frac{0.00015 \, M \times 0.00015 \, M}{0.0001602 \, M} \right)

pH=−log⁡(0.0000225)pH = -\log \left( 0.0000225 \right)

pH=3.65pH = 3.65

Conclusion

In this article, we have explored the titration of a weak base with a strong acid, specifically the titration of 30.0 mL of 0.267 M weak base $B \left( K_b =1.3 \times 10^{-10}\right)$ with 0.150 M HI. We have calculated the pH of the solution after the addition of 20.0 mL of HI, resulting in a pH of 3.65.

References

  • Atkins, P. W., & De Paula, J. (2010). Physical chemistry. Oxford University Press.
  • Brown, T. E., & LeMay, H. E. (2011). Chemistry: The Central Science. Pearson Education.
  • Chang, R. (2010). Chemistry: The Science in Context. McGraw-Hill.
    Titration of a Weak Base with a Strong Acid: Q&A =====================================================

Introduction

In our previous article, we explored the titration of a weak base with a strong acid, specifically the titration of 30.0 mL of 0.267 M weak base $B \left( K_b =1.3 \times 10^{-10}\right)$ with 0.150 M HI. We calculated the pH of the solution after the addition of 20.0 mL of HI, resulting in a pH of 3.65. In this article, we will answer some frequently asked questions related to the titration of a weak base with a strong acid.

Q: What is the difference between a weak base and a strong base?

A: A weak base is a base that does not completely dissociate in water, resulting in a solution with a pH greater than 7. A strong base, on the other hand, completely dissociates in water, resulting in a solution with a pH greater than 7.

Q: What is the difference between a strong acid and a weak acid?

A: A strong acid is an acid that completely dissociates in water, resulting in a solution with a pH less than 7. A weak acid, on the other hand, does not completely dissociate in water, resulting in a solution with a pH greater than 7.

Q: What is the purpose of titration in chemistry?

A: The purpose of titration in chemistry is to determine the concentration of a substance by reacting it with a known amount of another substance.

Q: How do you calculate the pH of a solution after the addition of a strong acid to a weak base?

A: To calculate the pH of a solution after the addition of a strong acid to a weak base, you need to consider the equilibrium between the weak base and the strong acid. You can use the formula:

pH=−log⁡([BH+][I−][B])pH = -\log \left( \frac{[BH^+][I^-]}{[B]} \right)

where $pH$ is the pH of the solution, $[BH^+]$ is the concentration of the conjugate acid, $[I^-]$ is the concentration of the conjugate base, and $[B]$ is the concentration of the weak base.

Q: What is the significance of the base dissociation constant (Kb) in the titration of a weak base with a strong acid?

A: The base dissociation constant (Kb) is a measure of the strength of a base. A higher Kb value indicates a stronger base, while a lower Kb value indicates a weaker base. In the titration of a weak base with a strong acid, the Kb value is used to calculate the pH of the solution.

Q: Can you give an example of a weak base and a strong acid?

A: Yes, an example of a weak base is ammonia (NH3), which has a Kb value of 1.8 x 10^-5. An example of a strong acid is hydrochloric acid (HCl), which completely dissociates in water.

Q: What is the difference between a titration curve and a pH curve?

A: A titration curve is a graph that shows the pH of a solution as a function of the volume of a strong acid added to a weak base. A pH curve, on the other hand, is a graph that shows the pH of a solution as a function of the concentration of a substance.

Conclusion

In this article, we have answered some frequently asked questions related to the titration of a weak base with a strong acid. We hope that this article has provided you with a better understanding of the titration process and the factors that affect the pH of a solution.

References

  • Atkins, P. W., & De Paula, J. (2010). Physical chemistry. Oxford University Press.
  • Brown, T. E., & LeMay, H. E. (2011). Chemistry: The Central Science. Pearson Education.
  • Chang, R. (2010). Chemistry: The Science in Context. McGraw-Hill.