Presentation on theme: "Section 16.3 Titrations and Buffers 1.To know how to neutralize acids and bases (titration) Learning Goal."— Presentation transcript:
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 -