What is the difference between a diprotic and triprotic acid? How does each acid ionise in water?
proton: H+
di protic H2SO4, H2CO3
triprotic H3PO4
most ionize in steps
https://en.wikipedia.org/wiki/Diprotic_acid
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A diprotic acid is an acid that can donate two protons (H+) per molecule when dissolved in water, while a triprotic acid is an acid that can donate three protons (H+) per molecule when dissolved in water. The main difference between them is the number of protons they can release.
To understand how these acids ionize in water, it's important to know that when an acid dissolves in water, it tends to dissociate into its constituent ions. Let's take a look at each acid:
1. Diprotic acid (e.g., sulfuric acid, H2SO4):
- In the first ionization step, H2SO4 donates one proton (H+) to water and becomes HSO4-. This step can be represented as: H2SO4 + H2O → HSO4- + H3O+
- In the second ionization step, the HSO4- ion donates another proton (H+), becoming SO4^2-. This step can be represented as: HSO4- + H2O → SO4^2- + H3O+
2. Triprotic acid (e.g., phosphoric acid, H3PO4):
- In the first ionization step, H3PO4 donates one proton (H+) to water and becomes H2PO4-. This step can be represented as: H3PO4 + H2O → H2PO4- + H3O+
- In the second ionization step, the H2PO4- ion donates another proton (H+), becoming HPO4^2-. This step can be represented as: H2PO4- + H2O → HPO4^2- + H3O+
- In the third ionization step, the HPO4^2- ion donates another proton (H+), becoming PO4^3-. This step can be represented as: HPO4^2- + H2O → PO4^3- + H3O+
In both cases, the acids gradually lose protons as they interact with water. The resulting hydronium ions (H3O+) combine with the remaining acid anions to form the final ions.
It's worth noting that as the acidity level increases (from diprotic to triprotic acids), each subsequent ionization step becomes weaker due to the loss of positive charge on the acid molecule. Thus, each subsequent step will be less ionized compared to the previous step.