Titrations and pKa CHEMISTRY 1106

Titrations and Pka predictions  Titrations  Method of quantitative/chemical analysis which can be used to determine the concentration of a known reactant  SA/SB - HCl and NaOH  WA/SB - CH 3 COOH and NaOH  SA/WB – HCl and NH 4 OH  WA/WB - CH 3 COOH and NH 4 OH

Titrations 1 Titrations today will be SA/SB - HCl and NaOH SA/SB - HCl and NaOH WA/SB - CH 3 COOH and NaOH WA/SB - CH 3 COOH and NaOH Analyte Titrant Titrant – a reagent with known concentration (Usually the Base) Analyte – a solution in which the concentration is not known in advance (Usually Acid)

Equilibrium constants  Acid Dissociation Constant  Ka, is an equilibrium constant for the dissociation of an acid K acid = K a = [H 3 O + ] 1 [CH 3 COO - ] 1 [CH 3 COOH] 1

Ka  In this example we will be using a monoprotic acid HA H + + A - Ka = [H+][A-] [HA] pKa = -log Ka just like pH= -log [H+]  So how does pH and pKA relate?

1%/50%/99% Acid Dissociation  1% acid dissociation  [HA]/[A-] = 100 so, pH = pKa – 2  50% acid dissociation  [HA]/[A-] = 1 so, pH = pKa  99% acid dissociation  [HA]/[A-] = 0.01 so, pH = pKa + 2 On pKa scale Large pKa – indicates very weak acid Small pKa – indicates not such a weak acid

Poly-protic Acid (Oxalic Acid) 1EQ Pt. 2EQ Pt. pKa=pH

Ka calcultions K acid = K a = [H 3 O + ] 1 [CH 3 COO - ] 1 [CH 3 COOH] 1 [H 2 0] 1 Because [CH 3 COOH] = [CH 3 COO - ] then left with [H 3 O + ] So…………….. pH= -log [H 3 O + ] and –log Ka =-log[H 3 O + ] and pKa = -log [H 3 O + ] So………………. ½ Equivalence pH = pKa

pKa  Titrations  We usually measure the end point (visually)  Today we get the equivalence point (pH probe)  Use the data to find the greatest change and that is the eq. point  Take the equivalence point volume divide by 2 this is ½ the equivalence point ≈ pKa  CH 3 COOH + NaOH  H 2 O+ Na CH 3 COO  1 mole reacts with ½ mole NaOH  ½ mol CH 3 COOH and ½ mol CH 3 COO -

Base – Acid  pKb is related to the value of the pka and pkb via the equation of water  K w = [H + ] [OH − ]  And the equation

pkb Rearranging [OH] ion we get Substitution the value of the [OH] ion we get which reduces to

pkb  When K a, K b and K w are determined under the same conditions of temperature and ionic strength, it follows  Then pK b = pK w − pK a.  Thus doing a titration

Report  Is due after the all 4 titrations have been completed 3 times each  Each must have the pH equivalence point and the pKa value  Report and print all 4 graphs

Ph  Open microlab  Tab pH titrations  Open hand held titration program  Select “pH” under sensor  Bottom of sensor window open edit and read calibration curve “pH”  Finish  Ready to start

Running the program  Starting point on the burette does not matter.  Enter the volume you put in the burette  First measuremtne is 0 mL then is 5 mL then 10 ……..  Do no let the drops fall on the pH meter it alters the data

Last notes  Run the titration until a base plateau is reached stop if pH is above 11.5 for acid and 2.5 for the base.  Use NaOH to titrate the acid  Use HCl to titrate the base

Analysis  When done run a first derivative on the data using the analysis button  Choose plot derivative  Next window ok  Then derivative appears under the sensor and drag to spread sheet