using chemical equations, show how the triprotic acid H3PO4 ionizes. Phases are optional. Ka2<\/p>\n\n\n\n
The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n The ionization of phosphoric acid (H\u2083PO\u2084), a triprotic acid, occurs in three steps, each corresponding to the loss of a proton (H\u207a) from the molecule. The ionization can be represented by the following chemical equations:<\/p>\n\n\n\n First ionization:<\/strong> Second ionization:<\/strong> Third ionization:<\/strong> Each ionization step is characterized by its corresponding acid dissociation constant (Ka). The Ka values for each step decrease with each successive ionization, meaning that the first ionization is the strongest, and the third ionization is the weakest.<\/p>\n\n\n\n The first dissociation (Ka\u2081) represents the loss of a proton from H\u2083PO\u2084 to form H\u2082PO\u2084\u207b. This reaction is relatively strong, with Ka\u2081 being about 7.5\u00d710\u221237.5 \\times 10^{-3}. The second dissociation (Ka\u2082) involves the loss of a proton from H\u2082PO\u2084\u207b to form HPO\u2084\u00b2\u207b. The value of Ka\u2082 is smaller, approximately 6.2\u00d710\u221286.2 \\times 10^{-8}, showing that this ionization is weaker than the first. The third dissociation (Ka\u2083) is even weaker, with Ka\u2083 being around 4.8\u00d710\u2212134.8 \\times 10^{-13}, indicating that HPO\u2084\u00b2\u207b loses its proton to form PO\u2084\u00b3\u207b only under very specific conditions.<\/p>\n\n\n\n The successive decrease in Ka values shows that each proton is more difficult to lose than the previous one. The Ka values also reflect the increasing stability of the ions formed in each step. The process of ionization in H\u2083PO\u2084 is essential in understanding its role as a buffer in biological systems, where it helps maintain pH balance by accepting or donating protons depending on the conditions.<\/p>\n\n\n\n
H3PO4(aq)\u21ccH+(aq)+H2PO4\u2212(aq)H_3PO_4 (aq) \\rightleftharpoons H^+ (aq) + H_2PO_4^- (aq)<\/p>\n\n\n\n
H2PO4\u2212(aq)\u21ccH+(aq)+HPO42\u2212(aq)H_2PO_4^- (aq) \\rightleftharpoons H^+ (aq) + HPO_4^{2-} (aq)<\/p>\n\n\n\n
HPO42\u2212(aq)\u21ccH+(aq)+PO43\u2212(aq)HPO_4^{2-} (aq) \\rightleftharpoons H^+ (aq) + PO_4^{3-} (aq)<\/p>\n\n\n\n