Chem 103 Lecture 3B Today: Acids and Bases5

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1. Calculating pH in 2 more scenarios: a) Pure weak base b) Buffer solutions Chem 103 Lecture 3b Today: Acids and Bases5

1. Acid base titration 2. Polyprotic acids

Ch 103 Seating arrangement Neutralization reactions

In preparation for group work and midterm exams: Seating arrangement enforced for extra credit work. When an acid encounters a base, neutralization occurs: (if in wrong row, no extra credit) Generic case: Strong acid + strong base ---> salt + water

If your last name starts with please sit in row Example: HCl(aq) + NaOH(aq) --> NaCl(aq) + H2O(l) Ionic eq.: H+ + Cl- + Na+ + OH- --> Na+ + Cl- + H O A, B or C A (front) 2 Important is net ionic eq: H+ + OH- --> H O (spectators: Na+,Cl-) D, E, F, G, H, I, J B 2 Example: HA(aq) + NaOH(aq) --> NaA(aq) + H2O K, L, M C - - Net ionic eq: HA(aq) + OH (aq) --> A (aq) + H2O(l) N, O, P, Q, R, Sa D Strong acid + strong base ---> 100% completion Si, T, U, V, W, X, Y, Z E Strong + weak ---> 100% completion Weak + weak ---> not 100% completion

Ways a buffer can result ABC’s of titration A=acid, B=base, C=calculations

When you add HCl to NH3 what is the net ionic equation? Titrant

Chemical rxn: HCl (aq) + NH3 (aq) ---> NH4Cl + - + - Ionic eq.: H + Cl + NH3 ---> NH4 + Cl + + Net ionic equation: H + NH3 --> NH4 + If you add 0.50 moles of H to 1.00 moles of NH3 what remains? + Answer: 0.50 moles of NH4 and 0.50 moles of NH3 I.e. A buffer containing base and its conjugate acid. Analyte of known volume + pH indicator

Ultimate purpose: to determine molarity of analyte Acid base titration Acid base titration, a quicker version In titration, determine the equivalence point (ep). Look at the same 1:1 titration problem

Simple case: HCl + NaOH --> H2O + NaCl ( a 1:1 titration) Titration of 20.0 mLs of NaOH requires 15.0 mLs of 0.120 M Titration of 20.0 mLs of NaOH requires 15.0 mLs of 0.120 M hydrochloric acid, HCl, to reach the equivalence point (ep). hydrochloric acid, HCl, to reach the equivalence point (ep). What is the concentration of the NaOH analyte? What is the concentration of the NaOH analyte? Solution: chem. rxn:HCl + NaOH --> H O + NaCl HCl + NaOH --> H2O + NaCl Solution: 1 mol NaOH 2 #mol HCl x [NaOH] = #mol NaOH = 1 mol HCl At equivalence pt (ep): # equiv HCl = # equiv NaOH L L But # equiv HCl = # mol HCl x 1 H+/mole = # mol HCl 1 mol NaOH MHClVHCl x Same for NaOH so: # mol HCl = #mol NaOH But #mol HCl = MHClVHCl so: 1 mol HCl L So M V = M V => “ M V = M V “ ! ! HCl HCl NaOH NaOH 1 1 2 2 (0.120M)(0.0150 L)(1/1) M2 = M1V1 / V2 = (0.120M)(15.0mL)/(20.0mL) ﬁnally: [NaOH] = 0.0200L = 0.0900M M2 = 0.0900M !

Acid base titration Continuation…

In titration, determine the equivalence point (ep). In previous titration example: VMg(OH)2 = 20.0 mL, MH3PO4 = 0.120M, Ve=15.0 mL

Example: 2 H3PO4 + 3 Mg(OH)2 --> 6 H2O + Mg3(PO4)2 2 H3PO4 + 3 Mg(OH)2 --> 6 H2O + Mg3(PO4)2

Titration of 20.0 mLs of Mg(OH)2 requires 15.0 mLs of 0.120 Solution: M phosphoric acid, H PO , to reach the equiv. pt (ep). What mol H PO 3 mol Mg(OH) 3 4 [Mg(OH) ] = mol Mg(OH) 2 = 3 4 x 2 is the concentration of the Mg(OH) analyte? 2 2 L L 2 mol H3PO4

2 H3PO4 + 3 Mg(OH)2 --> 6 H2O + Mg3(PO4)2 = MH3PO4VH3PO4 3 mol Mg(OH)2 = (0.120M)(15.0mL) 3 Solution: x x ! L 2 mol! H3PO4 20.0mL 2 mol Mg(OH) mol H PO 3 mol Mg(OH) [Mg(OH)2] = 2 = 3 4 x 2 L L 2 mol H3PO4 = 0.135M ! !

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Using the other approach Strong Acid + strong base. pH = ?

In previous titration example: VMg(OH)2 = 20.0 mL, MH3PO4 = 0.120M, Ve=15.0 mL Problem:20.0 mLs of 1.00 M HCl + 21.0 mLs of 1.00 M NaOH. pH =?

2 H3PO4 + 3 Mg(OH)2 --> 6 H2O + Mg3(PO4)2 Solution: HCl + NaOH --> NaCl + H2O Solution: Always true at equivalenc point: NOTE: THIS IS NOT AT EQUIVALENCE! (Another type of problem!) (easiest is: “follow the moles”): # equivalents H3PO4 = # equivalents Mg(OH)2 mmol HCl=M V =(1.00M)(20.0mL)=20.0 mmol HCl but # eq = # mol x # H+ transferred: HCl HCl mmol NaOH=MNaOHVNaOH = (1.00M)(21.0mL)=21.0mmol NaOH So # mol H3PO4 x 3 = # mol Mg(OH)2 x 2 Limiting reagent is HCl ; mmol NaOH excess=21.0-20.0=1.0mmol

3 M1V1 = 2 M2V2 where subscript “1” = H3PO4 ; “2” = Mg(OH)2 So [OH-] = mole OH-/total Vol(L) = mmol OH-/total mLs 3M1V1 3(0.120)(15.0) [OH-]=1.0mmol/(20.0+21.0)mL= 1.0/41.0 M = 0.0244 M So: M2 = = M = 0.135 M 2V2 2 (20.0) pOH = -log(0.0244)= 1.613; pH=14.00-1.653= 12.38

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