Question

What is the equilibrium associated with ka3 for h3po4?

Answer 1

The equilibrium associated with $\text{K}_{a3}$ for $\text{H}_3\text{PO}_4$ is as follows:

 $\boxed{\text{HPO}^{2-}_{4}(aq)+\text{H}_{2}\text{O}(l)_{\rightleftharpoons}^{{K}_{a3}}\text{PO}^{3-}_{4}(aq)+\text{H}_{3}\text{O}^{+}(aq)}$

Further Explanation:

Acids are substances that can donate hydrogen ion $\text{H}^{+}$ or proton in their aqueous solutions.

Classification of acids:

1. Monoprotic acids

Acids that are capable to donate single proton in solutions are monoprotic acids. For example, HCl is a monoprotic acid due to its tendency to donate only one proton in solutions.

2. Polyprotic acids

Acids that are capable to donate more than one proton in aqueous solutions are called polyprotic acids. For example, $\text{H}_2\text{SO}_4$ and $\text{H}_3\text{PO}_4$ are all polyprotic acids.

Acid strength is measured with the help of dissociation constant. Itis denoted by $\text{K}_a$. Since $\text{H}_3\text{PO}_4$ contains three hydrogens in it, it can donate three hydrogen ions in solutions so it is triprotic acid.

Dissociation of $\text{H}_3\text{PO}_4$ occurs as follows:

$\begin{aligned}\text{H}_3\text{PO}_4(aq)+\text{H}_2\text{O}(l)_{\rightleftharpoons}^{K_{a1}}\text{H}_2\text{PO}^{-}_{4}(aq)+\text{H}_3\text{O}^{+}(aq)\\\text{H}_2\text{PO}^{-}_{4}(aq)\text{H}_2\text{O}(l)_{\rightleftharpoons}^{K_{a3}}\text{HPO}^{2-}_{4}(aq)+\text{H}_{3}\text{O}^{+}(aq)\\\text{HPO}^{2-}_{4}(aq)+\text{H}_{2}\text{O}(l)_{\rightleftharpoons}^{K_{a3}}\text{PO}^{3-}_{4}(aq)+\text{H}_{3}\text{O}^{+}(aq)\end{aligned}$  

Where,

${{\text{k}}_{{\text{a1}}}}$ is the first dissociation constant of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$.

${{\text{k}}_{{\text{a2}}}}$ is the second dissociation constant of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$.

${{\text{k}}_{{\text{a3}}}}$ is the third dissociation constant of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$.

The expression of ${{\text{k}}_{{\text{a3}}}}$ for ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ is as follows:

${{\text{k}}_{{\text{a3}}}} = \dfrac{{\left[ {{\text{PO}}_4^{3 - }} \right]\left[ {{{\text{H}}_{\text{3}}}{{\text{O}}^ + }} \right]}}{{\left[ {{\text{HPO}}_4^{2 - }} \right]}}$  

Learn more:

1. Calculation of equilibrium constant of pure water at 25°C: brainly.com/question/3467841
2. Complete equation for the dissociation of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ (aq): brainly.com/question/5425813

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Equilibrium

Keywords: acids, monoprotic acid, polyprotic acid, hydrogen ions, ka3, H3PO4, acid strength, dissociation constant, first dissociation constant, second dissociation constant, third dissociation constant.

Answer 2

Phosphoric acid has 3 pKa values (pKa1=2.1,  pKa2=6.9, pKa3= 12.4) and after 3 ionization it gives 3 types of ions at different pKa values: 

H₃PO₄(aq) + H₂O(l) ⇌ H₃O⁺(aq) + H₂PO₄⁻ (aq)         pKₐ₁ 


H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HPO₄²⁻ (aq)       pKₐ₂


HPO₄²⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + PO₄³⁻ (aq)          pKₐ₃ 

The last equilibrium is associated with the highest pKa value (12.4) of phosphoric acid. There the last OH group will lose its hydrogen and hydrogen phosphate ion (HPO₄²⁻) turns into phosphate ion (PO₄³⁻).