What is the pH of 0.26 M H₃PO₄?
The Ka values for phosphoric acid are:
Ka1=7.5×10−3
Ka2=6.2×10−8
Ka3=3.6×10−13
The Correct Answer and Explanation is:
To calculate the pH of a 0.26 M solution of phosphoric acid (H₃PO₄), we need to consider its dissociation in water and the respective acid dissociation constants (Ka values). Phosphoric acid is a triprotic acid, meaning it can donate three protons (H⁺ ions). The dissociation steps are as follows:
- First Dissociation (Ka1):
[
\text{H}3\text{PO}_4 \rightleftharpoons \text{H}^+ + \text{H}_2\text{PO}_4^- ] [ K{a1} = 7.5 \times 10^{-3}
] - Second Dissociation (Ka2):
[
\text{H}2\text{PO}_4^- \rightleftharpoons \text{H}^+ + \text{HPO}_4^{2-} ] [ K{a2} = 6.2 \times 10^{-8}
] - Third Dissociation (Ka3):
[
\text{HPO}4^{2-} \rightleftharpoons \text{H}^+ + \text{PO}_4^{3-} ] [ K{a3} = 3.6 \times 10^{-13}
]
Given that (Ka1) is significantly larger than (Ka2) and (Ka3), we can assume that the first dissociation step will contribute the most to the concentration of H⁺ ions in solution.
Step 1: Calculate ([H^+]) from the first dissociation.
Using the expression for (Ka1):
[
K_{a1} = \frac{[\text{H}^+][\text{H}2\text{PO}_4^-]}{[\text{H}_3\text{PO}_4]} ] Assuming (x) is the amount that dissociates: [ K{a1} = \frac{x \cdot x}{0.26 – x} \approx \frac{x^2}{0.26}
]
Setting (K_{a1} = 7.5 \times 10^{-3}), we get:
[
7.5 \times 10^{-3} = \frac{x^2}{0.26}
]
[
x^2 = 7.5 \times 10^{-3} \times 0.26 \approx 1.95 \times 10^{-3}
]
[
x \approx \sqrt{1.95 \times 10^{-3}} \approx 0.0443 \text{ M}
]
Step 2: Calculate pH.
Since (x) represents ([H^+]):
[
\text{pH} = -\log([H^+]) = -\log(0.0443) \approx 1.35
]
Conclusion: The pH of a 0.26 M solution of phosphoric acid (H₃PO₄) is approximately 1.35. The calculation demonstrates the predominance of the first dissociation step, which has the largest (K_a), while the contributions from subsequent dissociations are negligible due to their much smaller (K_a) values.