CHEM 1332 1st Edition Lecture 17 Outline of Last Lecture - Anions- Cations- Polyprotic AcidsOutline of Current Lecture - Molarity- Spectator ions- Neutralization- Solving pH equationsCurrent Lecture- To distinguish between acids and bases we look at the + ¿H3O¿¿−¿O H¿¿- There are four steps that need to be taken to solve pH equations- First step is to find the molarity of reactants if it is not given to us molarity is moles of solute divided by liters of solution- If you are given 0.1 mol of HCl in I L solution you use the molarity equation find that the molarity is 0.1 M so you then find that the pH is 1- The second step is to eliminate spectator ions for cations and there are seven+¿+¿ , K¿+¿ , N a¿+¿ , L i¿+¿ , R b¿2+¿ , C r¿+¿ , B a¿Sr¿¿ for anions there are five which are −¿−¿ ,Cl O4¿−¿ , N O3¿−¿ , I¿−¿ , B r¿C l¿¿- The third step is to neutralize if there is an acid and base in the problemThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- You start with −¿ →−¿+¿+O H¿H¿- Next weak acid +−¿ →O H¿Conjugate base - Next weak base ++¿ →H¿Conjugate acid- This last two steps represent the change or limiting reagent- For example if you are given 0.1 +¿H¿and 0.2 −¿O H¿ the limiting reagent is +¿H¿so there is 0.1 M of −¿O H¿left so you solve and the pH is 13- To neutralize acid +basesalt +water-+¿H¿anion + cation−¿O H¿yields ionic compound- There are three types of neutralization- The first type is strong acid plus strong base and it goes to completion - The second type is weak acid plus strong base and it yields a conjugate base- The third type is a strong acid and weak base and it yields a conjugate acid- The forth step to solving the pH equation is looking at what you are left with and decide the steps that are needed to solve for the pH- If you are given +¿H¿ you go from here to solve for the pH- If you are given −¿O H¿you solve for pOH and from this get the pH- If you are given −¿HA+ H2O ⇄ H3O+ A¿ Use the Ka value to get the +¿H¿ and then the pH- If you are given B−¿+¿+O H¿+ H2O ⇄ B H¿Use the Kb value to get −¿O H¿then pOH then pH- If you are given − ¿+ HA−¿ H2O⇄ O H¿A¿ Use Kb value to get −¿O H¿then pOH then pH- If you are given BH + H2O ⇄ H3O+ B Use Ka to get the +¿H¿ and then the pHKey TermsSpectator ions- ions that do not affect