Question
calculate pH of 0.01 M sulfuric acid solution
Answers
Answered by
GPT-4o mini
Sulfuric acid (H₂SO₄) is a strong acid and dissociates completely in water. The first dissociation step is:
\[ \text{H}_2\text{SO}_4 \rightarrow \text{H}^+ + \text{HSO}_4^- \]
In a 0.01 M sulfuric acid solution, the first dissociation yields 0.01 M of \( \text{H}^+ \) ions from the H₂SO₄.
The second dissociation of the bisulfate ion (\( \text{HSO}_4^- \)) can also contribute to the concentration of \( \text{H}^+ \) ions:
\[ \text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-} \]
This second dissociation is a bit weaker than the first, but for simplicity, we can estimate the contribution of \( \text{H}^+ \) from this second dissociation. Typically, at a concentration of 0.01 M, the contribution from the second dissociation is relatively small, but can still be roughly estimated to be around 0.005 M.
So, the total concentration of \( \text{H}^+ \) ions can be approximated as follows:
\[ [\text{H}^+] \approx 0.01 \, \text{M} + 0.005 \, \text{M} = 0.015 \, \text{M} \]
To calculate the pH:
\[
\text{pH} = -\log[\text{H}^+]
\]
Substituting the total concentration of hydrogen ions:
\[
\text{pH} \approx -\log(0.015)
\]
Using a calculator, we find:
\[
\text{pH} \approx 1.83
\]
Thus, the pH of a 0.01 M sulfuric acid solution is approximately **1.83**.
\[ \text{H}_2\text{SO}_4 \rightarrow \text{H}^+ + \text{HSO}_4^- \]
In a 0.01 M sulfuric acid solution, the first dissociation yields 0.01 M of \( \text{H}^+ \) ions from the H₂SO₄.
The second dissociation of the bisulfate ion (\( \text{HSO}_4^- \)) can also contribute to the concentration of \( \text{H}^+ \) ions:
\[ \text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-} \]
This second dissociation is a bit weaker than the first, but for simplicity, we can estimate the contribution of \( \text{H}^+ \) from this second dissociation. Typically, at a concentration of 0.01 M, the contribution from the second dissociation is relatively small, but can still be roughly estimated to be around 0.005 M.
So, the total concentration of \( \text{H}^+ \) ions can be approximated as follows:
\[ [\text{H}^+] \approx 0.01 \, \text{M} + 0.005 \, \text{M} = 0.015 \, \text{M} \]
To calculate the pH:
\[
\text{pH} = -\log[\text{H}^+]
\]
Substituting the total concentration of hydrogen ions:
\[
\text{pH} \approx -\log(0.015)
\]
Using a calculator, we find:
\[
\text{pH} \approx 1.83
\]
Thus, the pH of a 0.01 M sulfuric acid solution is approximately **1.83**.