What Is The PH Of A 0.335 M Solution Of $H_2SO_4$?Given: $K_{a2} = 1.20 \times 10^{-2}$

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Introduction

Sulfuric acid ($H_2SO_4$) is a strong acid that completely dissociates in water to produce hydrogen ions ($H^+$) and sulfate ions ($SO_4^{2-}$). However, the dissociation of sulfuric acid is a two-step process, with the first step being the dissociation of the first hydrogen ion and the second step being the dissociation of the second hydrogen ion. The dissociation constant for the second step, $K_{a2}$, is given as $1.20 \times 10^{-2}$.

Understanding the Dissociation of Sulfuric Acid

The dissociation of sulfuric acid can be represented by the following equation:

H2SO4⇌H++HSO4−H_2SO_4 \rightleftharpoons H^+ + HSO_4^-

This is the first step in the dissociation process, and it has a very large dissociation constant ($K_a$) of approximately $10^{6}$, indicating that it is a strong acid and completely dissociates in water.

The second step in the dissociation process is the dissociation of the hydrogen sulfate ion ($HSO_4^-$) to produce another hydrogen ion and a sulfate ion:

HSO4−⇌H++SO42−HSO_4^- \rightleftharpoons H^+ + SO_4^{2-}

This step has a dissociation constant ($K_{a2}$) of $1.20 \times 10^{-2}$, which is much smaller than the first step.

Calculating the pH of the Solution

To calculate the pH of the solution, we need to consider the second step in the dissociation process, as the first step is complete and does not affect the pH of the solution.

The dissociation constant for the second step ($K_{a2}$) is given as $1.20 \times 10^{-2}$, and the initial concentration of sulfuric acid ($H_2SO_4$) is 0.335 M.

We can use the following equation to calculate the concentration of hydrogen ions ($H^+$) produced in the second step:

Ka2=[H+][SO42−][HSO4−]K_{a2} = \frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

Since the dissociation of sulfuric acid is a two-step process, we need to consider the concentration of hydrogen sulfate ions ($HSO_4^-$) produced in the first step.

The concentration of hydrogen sulfate ions ($HSO_4^-$) can be calculated using the following equation:

[HSO4−]=[H2SO4]×KaKa+Ka2[HSO_4^-] = [H_2SO_4] \times \frac{K_a}{K_a + K_{a2}}

Substituting the values, we get:

[HSO4−]=0.335×106106+1.20×10−2[HSO_4^-] = 0.335 \times \frac{10^{6}}{10^{6} + 1.20 \times 10^{-2}}

Simplifying the equation, we get:

[HSO4−]=0.335×106106+1.20×10−2≈0.335[HSO_4^-] = 0.335 \times \frac{10^{6}}{10^{6} + 1.20 \times 10^{-2}} \approx 0.335

Now, we can substitute the values into the equation for $K_{a2}$:

1.20×10−2=[H+][SO42−][HSO4−]1.20 \times 10^{-2} = \frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

Substituting the values, we get:

1.20×10−2=[H+][0.335]0.3351.20 \times 10^{-2} = \frac{[H^+][0.335]}{0.335}

Simplifying the equation, we get:

[H+]=1.20×10−2×0.335[H^+] = \sqrt{1.20 \times 10^{-2} \times 0.335}

Solving for $[H^+]$, we get:

[H+]≈0.011[H^+] \approx 0.011

Calculating the pH of the Solution

Now that we have the concentration of hydrogen ions ($H^+$), we can calculate the pH of the solution using the following equation:

pH=−log⁡[H+]pH = -\log[H^+]

Substituting the value of $[H^+]$, we get:

pH=−log⁡(0.011)pH = -\log(0.011)

Solving for pH, we get:

pH≈1.96pH \approx 1.96

Conclusion

In this article, we calculated the pH of a 0.335 M solution of sulfuric acid ($H_2SO_4$) using the dissociation constant for the second step ($K_{a2}$) of $1.20 \times 10^{-2}$. We found that the pH of the solution is approximately 1.96.

References

  • Atkins, P. W., & De Paula, J. (2010). Physical chemistry. Oxford University Press.
  • Brown, T. E., & LeMay, H. E. (2012). Chemistry: The Central Science. Pearson Education.
  • Petrucci, R. H., Harwood, W. S., & Herring, F. G. (2011). General chemistry: Principles and modern applications. Pearson Education.

Introduction

In our previous article, we calculated the pH of a 0.335 M solution of sulfuric acid ($H_2SO_4$) using the dissociation constant for the second step ($K_{a2}$) of $1.20 \times 10^{-2}$. In this article, we will answer some frequently asked questions related to the calculation of pH.

Q: What is the significance of the dissociation constant ($K_{a2}$) in the calculation of pH?

A: The dissociation constant ($K_{a2}$) is a measure of the strength of the acid and is used to calculate the concentration of hydrogen ions ($H^+$) produced in the second step of the dissociation process.

Q: Why is the first step in the dissociation process not considered in the calculation of pH?

A: The first step in the dissociation process is complete and does not affect the pH of the solution. The second step is the rate-determining step, and its dissociation constant ($K_{a2}$) is used to calculate the pH of the solution.

Q: What is the relationship between the concentration of hydrogen sulfate ions ($HSO_4^-$) and the pH of the solution?

A: The concentration of hydrogen sulfate ions ($HSO_4^-$) is directly related to the pH of the solution. As the concentration of hydrogen sulfate ions ($HSO_4^-$) increases, the pH of the solution decreases.

Q: How does the concentration of sulfuric acid ($H_2SO_4$) affect the pH of the solution?

A: The concentration of sulfuric acid ($H_2SO_4$) affects the pH of the solution by increasing the concentration of hydrogen ions ($H^+$) produced in the second step of the dissociation process.

Q: What is the effect of temperature on the dissociation constant ($K_{a2}$)?

A: The dissociation constant ($K_{a2}$) is temperature-dependent and increases with increasing temperature.

Q: How does the pH of the solution change with increasing concentration of sulfuric acid ($H_2SO_4$)?

A: The pH of the solution decreases with increasing concentration of sulfuric acid ($H_2SO_4$).

Q: What is the relationship between the pH of the solution and the concentration of hydrogen sulfate ions ($HSO_4^-$)?

A: The pH of the solution is inversely related to the concentration of hydrogen sulfate ions ($HSO_4^-$).

Q: How does the dissociation constant ($K_{a2}$) affect the pH of the solution?

A: The dissociation constant ($K_{a2}$) affects the pH of the solution by increasing the concentration of hydrogen ions ($H^+$) produced in the second step of the dissociation process.

Conclusion

In this article, we have answered some frequently asked questions related to the calculation of pH of a 0.335 M solution of sulfuric acid ($H_2SO_4$). We hope that this article has provided a better understanding of the relationship between the dissociation constant ($K_{a2}$) and the pH of the solution.

References

  • Atkins, P. W., & De Paula, J. (2010). Physical chemistry. Oxford University Press.
  • Brown, T. E., & LeMay, H. E. (2012). Chemistry: The Central Science. Pearson Education.
  • Petrucci, R. H., Harwood, W. S., & Herring, F. G. (2011). General chemistry: Principles and modern applications. Pearson Education.