synthetic ion exchange resins. These resins are organic polymeric networks that contain functional groups that are permanently attached to the resin. One such functional group is sulfonic acid, -SO3H. When placed in water, the resin (being a large organic molecule) will not dissolve. However, the ionic functional groups become hydrated and the resin will swell as it takes up waters of hydration. The resin may then be thought of as a solid solution mixture with the functional groups being "dissolved" but the polymer remaining as a solid. The hydronium ion (H3O+) associated with acidic functional groups like sulfonic acid will dissociate just as it would if the acid were free of the resin. If other cations such as Cu2+ or Na+ are added to the resin, an ion-exchange reaction can occur: . Note that one H+ is released for each positive charge bound to the resin. Since the resin exchanges a hydrogen ion for a sodium ion in this reaction, sulfonic acid resins are called cation-exchange resins. Equilibrium constants may be written for such reactions. For example, K= . For most cations the value of this K is large. Thus, the addition of a cationic material to the resin will displace an equivalent amount of H+ ions. If concentrated acid is added to a resin that is bound by metal ions, the equilibrium will shift to the left and the metal ions will be freed and the resin will return to the acid (H+) form. In this experiment, volumes of NaCl, Fe(NO3)3, and NaOH will be passed through a column packed with resin in the H+ form. The H+ released from the resin by each cation (Na+ and Fe3+) will be measured by titration with NaOH. Question explain why the endpoint was reached so quickly for the NaOH samples
synthetic ion exchange resins. These resins are organic
. Note that one H+ is released for each positive charge bound to the resin.
Since the resin exchanges a hydrogen ion for a sodium ion in this reaction, sulfonic acid resins are called cation-exchange resins. Equilibrium constants may be written for such reactions. For example, K= . For most cations the value of this K is large. Thus, the addition of a cationic material to the resin will displace an equivalent amount of H+ ions. If concentrated acid is added to a resin that is bound by metal ions, the equilibrium will shift to the left and the metal ions will be freed and the resin will return to the acid (H+) form. In this experiment, volumes of NaCl, Fe(NO3)3, and NaOH will be passed through a column packed with resin in the H+ form. The H+ released from the resin by each cation (Na+ and Fe3+) will be measured by titration with NaOH.
Question
explain why the
endpoint was reached so quickly for the NaOH samples.
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