Understanding the pH of 0.05 M H3PO4

Introduction to H3PO4

Phosphoric acid (H3PO4) is a triprotic weak acid, meaning it can donate three protons (H ). This makes its pH calculation more complex compared to strong acids like hydrochloric acid. Understanding the pH value of 0.05 M H3PO4 involves a step-by-step approach to its ionization and dissociation.

Assumptions and Clarification

The question initially provided confusing information, suggesting that H3PO4 is completely ionized or a strong acid, which is incorrect. H3PO4 is a weak acid, and the assumption of complete ionization is a significant oversimplification. Here, we are calculating the pH of 0.05 M H3PO4 under the assumption of its weak acidic behavior.

Dissociation Constants of H3PO4

The first step involves understanding the dissociation constants of phosphoric acid: - K1 6.910^-3 - K2 6.310^-8 - K3 4.810^-13 Since K1 is much larger than K2 and K3, the major source of H ions comes from the first dissociation stage. Common ion effect from the first dissociation will significantly reduce the second and third dissociation.

Calculation of pH

Let's begin with the first dissociation of H3PO4: [ H_3PO_4 rightarrow H^ H_2PO_4^- ] Using the first dissociation constant, K1, we have: [ K1 frac{[H^ ][H_2PO_4^-]}{[H_3PO_4]} ] Given the initial concentration of H3PO4 is 0.05 M, we can set up the following equation: [ 6.9 times 10^{-3} frac{x^2}{0.05 - x} ] Neglecting x in the denominator, we get: [ 6.9 times 10^{-3} times 0.05 x^2 ] [ x sqrt{6.9 times 10^{-4}} 0.026 text{ M} ] Thus, [ [H^ ] x 0.026 text{ M} ] And, [ pH -log[0.026] 1.58 ]

Refinement Using Successive Approximations

To refine the calculation, we consider: [ frac{[H^ ][H_2PO_4^-]}{[H_3PO_4]} frac{x^2}{0.05 - x} 6.9 times 10^{-3} ] Assuming [H_2PO_4^-] [H^ ] x, [ frac{x^2}{0.05 - x} 6.9 times 10^{-3} ] Solving this equation, we get: [ x 0.025 text{ M} ] Thus, [ [H^ ] 0.025 text{ M} ] And, [ pH -log[0.025] 1.60 ]

Further Impact of Second Dissociation

For the second dissociation, [ H_2PO_4^- rightarrow H^ HPO_4^{2-} ] Using K2, [ K2 frac{[H^ ][HPO_4^{2-}]}{[H_2PO_4^-]} frac{x[H^ ]}{0.025 - x} ] Assuming [H_2PO_4^-] approx 0.025 M and [HPO_4^{2-}] approx K2/[H^ ] 6.3 times 10^{-8} [H^ ], [ [H^ ] approx 0.025 6.3 times 10^{-8} [H^ ] ] Thus, the contribution of the second dissociation is negligible, and we can confidently use the first dissociation for pH calculation.

Conclusion

Given the detailed step-by-step approach, the pH of 0.05 M H3PO4 is approximately 1.60, confirming the significant contribution of the first dissociation in determining the pH of the solution. This understanding is crucial for similar problems involving weak acids and their ionization.

This article provides a comprehensive explanation of the pH calculation of 0.05 M H3PO4. Understanding the behavior of weak acids like H3PO4 is essential in various scientific and industrial applications. You can apply these principles to calculate the pH of solutions involving weak acids and troubleshoot similar problems.