# Section 16.3 Titrations and Buffers 1.To know how to neutralize acids and bases (titration) Learning Goal.

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Section 16.3 Titrations and Buffers 1.To know how to neutralize acids and bases (titration) Learning Goal

Section 16.3 Titrations and Buffers Remember that… (add this to your notes) Acid + Base  Water + Salt Ex.: HCl + NaOH  H 2 O + NaCl Acid-Base Reactions (review) This is a double replacement reaction Notice that H + and OH -  H 2 O This is a neutralization reaction

Section 16.3 Titrations and Buffers Acid-Base Titrations Titration – delivering a measured volume of a solution of known concentration into a solution being analyzed  determine the concentration of the solution (analyte) We need to neutralize the solution, therefore… What should you add to an acid to neutralize it? What should you add to a base to neutralize it? BASE ACID

Section 16.3 Titrations and Buffers Titrant (standard solution) Analyte (solution being analyzed) The titration is stopped when the pH = 7 ↓ Buret – device used for accurate  measurement of the delivery of a liquid Stoichiometric point (equivalence point) – when just enough titrant has been added to react with all of the solution being analyzed A pH indicator is added to the analyte so the color changes when the pH reaches 7 (or a pH meter could be used) How will we know when the solution has been neutralized?

Section 16.3 Titrations and Buffers Acid-Base Titrations Titration curve (pH curve) – plot of the data (pH vs volume) for a titration  pH changes quickly close to the equivalence point.

Section 16.3 Titrations and Buffers Calculating the Volume of Titrant Needed in Acid-Base Titrations (using M=n/L) How much 0.100 M NaOH is needed to titrate 50.0 mL of 0.200 M HNO 3 ? 1.Calculate how many moles of H+ are in the analyte. 0.0500 L HNO 3 x 0.200 mol H + = 1.00 x 10 -2 mol H + 1L 2.Calculate the volume of titrant solution that contains the needed number of moles of OH - (same as moles of H + ). V = n/M = 1.00 x 10 -2 mol OH- = 1.00 x 10 -1 L 0.100 mol/L or 100. mL

Section 16.3 Titrations and Buffers Calculating Molarities and Volumes in Titrations (using one formula) Remember that Molarity x Volume (L) = n Moles of H + must equal moles of OH - for neutralization Since the acid or base may give off more than one H + or OH -, we must use the mole ratio of base/acid: M a V a (b/a) = M b V b Ex.: H 2 SO 4 + 2NaOH  2H 2 O + Na 2 SO 4 Notice that this is a double displacement/replacement reaction. In most neutralization reactions: acid + base  water + salt

Section 16.3 Titrations and Buffers To understand how the pH in a solution (eg. Blood) is kept constant when acid or base are added (buffers) Learning Goal

Section 16.3 Titrations and Buffers B. Buffered Solutions Buffered solution – resists a change in its pH when either and acid or a base has been added –Presence of a weak acid and its conjugate base buffers the solution: Buffer = weak acid + salt containing conj. base of acid

Section 16.3 Titrations and Buffers B. Buffered Solutions

Section 16.3 Titrations and Buffers Example of buffered solution: HC 2 H 3 O 2, NaC 2 H 3 O 2 weak acid salt that contains conjugate base of acid ( C 2 H 3 O 2 - ) If strong acid is added to the solution, the H + ions react with the conjugate base and form the weak acid: Net reaction: H + + C 2 H 3 O 2 -  HC 2 H 3 O 2 If strong base is added to the solution, the OH - ions extract the protons from the weak acid to form water: Net reaction: OH - + HC 2 H 3 O 2  H 2 O + C 2 H 3 O 2 -

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