Acids and Bases Titration Curves

Strong Acid-Base Titration Plotting Points 50.0 ml of [0.200] HNO3 is titrated with 0.100M NaOH. Calculate the pH points a. NaOH has NOT been added. Eq: HNO3 + NaOH  NaCl + NaNO3 Since HNO3 is a strong acid pH = 0.70 b. 10.0 ml of 0.100 M NaOH has been added. (or n=CV and 0.0010 moles) Acid moles in flask = C V = [0.200] x 0.0500L = 0.0100 mol to START H+ left over = 0.0100 mol Acid – 0.0010 mol OH- added = 0.0090mol Acid Left C Acid left = 0.0090mol H+/ 0.060 ml = 0.15 M Acid left so pH=0.82 c. Now another 10 ml base added for total of 20.0 mL of 0.100 M NaOH H+left over = 0.0100 mol – 0.0020 mol of OH- added = 0.0080mol Acid Left C Acid left = 0.0080mol H+/ 0.070 ml total volume = 0.114 M Acid left so pH=0.942 50.0 mL of 0.100 M NaOH has been added (half way). H+ = 0.0100 mol – 0.0050 mol of OH- added = 0.0050mol Acid left C Acid left = 0.0050mol H+/ 0.100 ml = 0.0500 M Acid left pH=1.13

Strong Acid (titrant) - Strong Base Titration (analyte) After 150.0 mL of 0.100 M NaOH has been added, Excess Base pH = 12.40 Center vertical region, Graph Midpoint = equivalence point, or look at products

Important points: pH increases slowly far from the equivalence point pH changes quickly near the equivalence point The equivalence point of a strong acid—strong base titration = 7.00 The titration of a strong base with a strong acid is almost identical, just reversed

COLOUR CHANGES OF SOME COMMON INDICATORS Acid-base indicators Must change immediately in the required pH range over the addition of ‘half’ a drop of reagent. COLOUR CHANGES OF SOME COMMON INDICATORS pH 1 2 3 4 5 6 7 8 9 10 11 12 13 14 METHYL ORANGE CHANGE LITMUS CHANGE PHENOLPHTHALEIN CHANGE

Titration Curves  A titration curve is a graph of the pH changes that occur during an acid-base titration versus the volume of acid or base added. You need to be able to recognize each and then choose a suitable indicator for that titration, by looking at the PRODUCTS or at the midpoint of the titration curves steepest section   Also at that very same point is the equivalence point, and the end of a titration where the stoichiometry exactly satisfied, or moles H+ = moles OH-. It is also called the transition point or end point referring to where the indicator changes color and [HInd] = [Ind-].

Choosing an Indicator The pH of the equivalence point depends on type of salt in PRODUCT.   Rule of thumb: Not exact pH’s but good approximations If the salt in PRODUCT is neutral the equivalence point = 7 HCl + K OH → K Cl + HOH If the salt in PRODUCT is basic the equivalence point = 9 CH3COOH + K OH -  H2O + K CH3COO - If the salt in PRODUCT is acidic the equivalence point = 5 HCl + NH3 → NH4+ + Cl-

Titration Curve: Strong Acid and Strong Base HCl + KOH → KCl + HOH   HCl + KOH → KCl + HOH 50 mL 50 mL of 0.10 M 0.10 M HCl Indicator pH = 7 Bromothymol Blue, why? LOOK AT THE PRODUCTS, neutral water and neutral salt, so pH=7, refer to your chart of indicators pH Volume .10 M KOH added 14 7 0.10 M KOH 0.10 M HCl Neutral Salt KCl 25 50

Titration Curve: Strong Acid (analyte)and Strong Base (titrant)   HCl + KOH → KCl + HOH Indicator pH = 7 Bromothymol Blue 50 mL of 0.10 M KOH is added to 50 mL of 0.10 M HCl pH Volume .10 M KOH added 14 7 25 50

NEXT Strong Base into a Weak Acid

Strong Base (titrant) into a Weak Acid (analyte) What is the end point and why? CH3COOH + OH -  H2O + CH3COO - Indicator pH = 9 Phenolphthalein, why? Weak base CH3COO - CH3COOH + OH -  H2O + CH3COO - Why is the pH higher at the start? It is due to the weak acid. Why? CH3COOH + OH -  H2O + CH3COO - 0.100 M of this weak acid is pH = 2.88

Strong Base (titrant) into a Weak Acid(analyte) Can you find the buffer area? CH3COOH + OH -  H2O + CH3COO - The titration curve levels off due to buffering effects. Buffering happens when [HA] become equal to [A-]. At exactly the halfway point to the equivalence point pH levels off near pKa due to HA/A- buffering effect pH = pKa + log([A-]/[HA]) = pKa + log(1) = pKa (when [A-] = [HA])

NEXT Weak Acid into a Strong Base

Titration Curve: Weak Acid into a Strong Base HCN + OH- → CN- + HOH 20 mL of 1.0 M HCN is added to 20 mL of 1.0 M KOH What is the end point and indicator and why? Indicator pH = 9 Phenolphthalein, why? The weak base CN- pH Volume 1.0 M HCN added 14 7 1.0 M KOH Weak base CN- 10 20

Titration Curve: Weak Acid into a Strong Base HCN + KOH → CN - + HOH Find the buffer region, why is it there? Buffer Zone excess HCN and CN - pH Volume 1.0 M HCN added 14 7 10 20

NEXT Weak BASE into a Strong Acid

Titration Curve: Weak Base into a Strong Acid HCl + NH3 → NH4+ + Cl- LAST Titration Curve: Weak Base into a Strong Acid   HCl + NH3 → NH4+ + Cl- 60 mL of 1.0 M NH3 is added to 60 mL of 1.0 M HCl Indicator pH = 5 Methyl Red, why? Weak acid in products NH4+ pH Volume 1.0 M NH3 added 14 7 1.0 M HCl 30 60

Titration Curve: Weak Base to a Strong Acid HCl + NH3 → NH4+ + Cl-   HCl + NH3 → NH4+ + Cl- Indicator pH = 5 Methyl Red pH Volume 1.0 M NH3 added 14 7 30 60 1.0 M NH3 1.0 M HCl Weak acid has little effect on Strong Base Acid Salt

Titration Curve: Strong Acid and Weak Base HCl + NH3 → NH4+ + Cl-   HCl + NH3 → NH4+ + Cl- Indicator pH = 5 Methyl Red 60 mL of 1.0 M NH3 is added to 60 mL of 1.0 M HCl pH Volume 1.0 M NH3 added 14 7 Buffer Zone NH3 and NH4+ 30 60

Next Weak with Weak Non-Stoichiometric

weak acid (CH3COOH) v. weak base (NH3) Curve levels off at pH 10 due to excess 0.1M NH3 (a weak alkali) NO SHARP CHANGE IN pH Steady pH change pH 4 due to 0.1M CH3COOH (weak monoprotic acid) Types

weak acid (CH3COOH) v. weak base (NH3) PHENOLPHTHALEIN LITMUS METHYL ORANGE NOTHING SUITABLE There is no suitable indicator- none change in the ‘vertical’ portion. The end point can be detected by plotting a curve using a pH meter.

POLYPROTIC ACID AND BASE TITRATION CURVES

Titration of a Weak Base with a Strong Acid Similar problem to the titration of a weak acid with a strong base Example: Titrate 100 ml of 0.10 M NH3 (Kb = 1.8 x 10-5) with 0.1 M HCl. Notice the weak acid get hit HARD and falls away quickly, but recovers due to buffer At ½ Equivalence point [HA] = [A-] thus pH = pKa  

pKa2 = 10.26 pKa1 = 6.36 Titrations of Polyprotic Acids and Bases Multiple Inflection Points = Multiple Equivalence Points will be seen CO32- + H+ HCO3- ½ way buffer point [HA]\[A-] = 1, Kb1 = KW/Ka2 = 1.8 x 10-4 (pKb1=3.74) HCO3- + H+ H2CO3 ½ way buffer point [HA]\[A-] = 1, Kb2 = KW/Ka1 = 2.3 x 10-8 (pKb2 =7.64) ½ Eq. pt 1 Eq. pt 1 ½ Eq. pt 2 Eq. pt 2 pKa2 = 10.26 (pKb1 = 3.74) pKa1 = 6.36 pKb2 = 7.64

Other pH curves - polyprotic acids (H3PO4) Phosphoric acid is triprotic; it reacts with sodium hydroxide in three steps... Step 1 H3PO4 + NaOH ——> NaH2PO4 + H2O Step 2 NaH2PO4 + NaOH ——> Na2HPO4 + H2O Step 3 Na2HPO4 + NaOH ——> Na3PO4 + H2O

Other pH curves - polyprotic acids (H3PO4) Phosphoric acid is triprotic; it reacts with sodium hydroxide in three steps... Step 1 H3PO4 + NaOH ——> NaH2PO4 + H2O Step 2 NaH2PO4 + NaOH ——> Na2HPO4 + H2O Step 3 Na2HPO4 + NaOH ——> Na3PO4 + H2O There are three sharp pH changes Each successive addition of NaOH is the same as equal number of moles are involved.

Other pH curves - polyprotic acids (H3PO4) Phosphoric acid is triprotic; it reacts with sodium hydroxide in three steps... Step 1 H3PO4 + NaOH ——> NaH2PO4 + H2O Step 2 NaH2PO4 + NaOH ——> Na2HPO4 + H2O Step 3 Na2HPO4 + NaOH ——> Na3PO4 + H2O pH of H3PO4 = 1.5

Other pH curves - polyprotic acids (H3PO4) Phosphoric acid is triprotic; it reacts with sodium hydroxide in three steps... Step 1 H3PO4 + NaOH ——> NaH2PO4 + H2O Step 2 NaH2PO4 + NaOH ——> Na2HPO4 + H2O Step 3 Na2HPO4 + NaOH ——> Na3PO4 + H2O pH of NaH2PO4 = 4.4 pH of H3PO4 = 1.5

Other pH curves - polyprotic acids (H3PO4) Phosphoric acid is triprotic; it reacts with sodium hydroxide in three steps... Step 1 H3PO4 + NaOH ——> NaH2PO4 + H2O Step 2 NaH2PO4 + NaOH ——> Na2HPO4 + H2O Step 3 Na2HPO4 + NaOH ——> Na3PO4 + H2O pH of Na2HPO4 = 9.6 pH of NaH2PO4 = 4.4 pH of H3PO4 = 1.5

Other pH curves - polyprotic acids (H3PO4) Phosphoric acid is triprotic; it reacts with sodium hydroxide in three steps... Step 1 H3PO4 + NaOH ——> NaH2PO4 + H2O Step 2 NaH2PO4 + NaOH ——> Na2HPO4 + H2O Step 3 Na2HPO4 + NaOH ——> Na3PO4 + H2O pH of Na3PO4 = 12 pH of Na2HPO4 = 9.6 pH of NaH2PO4 = 4.4 pH of H3PO4 = 1.5