Titration and pH Curves.

  A titration curve is a plot of pH vs. volume of added titrant.

Parts of a titration  Titrant- acid or base of known concentration that is added to the substance being analyzed.  Analyte- the substance that is being analyzed, or your unknown.

Titrations   Because titrations involve small concentrations, and mL are of often used in titrations, and millimoles, or mmol.   Molarity = mmol/mL   The equivalence point is when moles of titrant are equal to moles of analyte.   All volumes in a titration are considered to be additive.   Always label the equivalence point and for weak acid or base titrations the half equivalence point, where pH = pK a.

Strong Acid-Strong Base Titration Curves.  Before the addition.  pH is calculated directly from the initial concentration.  Additions before the equivalence point.  Construct a “stoichiometry” reaction table. Determine MOLES of acid in excess (not neutralized).  Divide MOLES by the TOTAL VOLUME to obtain [H 3 O + ].  Calculate the pH.

Strong Acid-Strong Base Titration Curves.   Additions at the equivalence point.   The pH ALWAYS is equal to 7.00 when [H 3 O + ] = [OH - ].   Additions beyond the equivalence point.   Construct a “stoichiometry” reaction table. Determine MOLES of base in excess (not neutralized).   Divide MOLES by the TOTAL VOLUME to obtain [OH - ].   Calculate the pOH, then the pH.

Problem   50.0 mL of M HNO 3 are titrated with M NaOH.   Calculate the pH after the additions of 0.0, 10, 20, 50, 100,150, and 200 mL samples of NaOH.   Then, construct a titration curve and label it properly.

Strong Acid Strong Base Problem   50.0 mL of 0.50 M HCl are titrated with M NaOH.   Calculate the pH after the additions of 0.0, 10, 20, 50, 100,150, and 200 mL samples of NaOH.   Then, construct a titration curve and label it properly.

Weak Acid-Strong Base Titration Curves.   Before the addition.   Construct an “equilibrium” reaction table ONLY!   K a = [A - ] [H 3 O + ] to obtain [H 3 O + ].   [HA]   Calculate the pH.

Weak Acid-Strong Base Titration Curves.   Additions before the equivalence point.   Construct a stoichiometry reaction table.   Determine MOLES of acid in excess (not neutralized) and MOLES of conjugate base formed.   Divide MOLES by the TOTAL VOLUME to obtain [H 3 O + ] and [A - ].   Construct an “equilibrium” reaction table.   K a = [A - ] [H 3 O + ] and obtain [H 3 O + ].   [HA]   Calculate the pH.

Half equivalence point  The equivalence point is when the moles or titrant are equal to the moles of analyte.  At half the equivalence point, exactly half of the weak analyte will be shifted into it’s conjugate.  HA ⇌ H + + A -, therefore [HA] = [A - ]  pH = pK a + log ([A - ]/[HA])  At the half equivalence point  pH = pK a

 Additions at the equivalence point.  Construct a stoichiometry reaction table.  Determine MOLES of conjugate base formed. Divide MOLES by the TOTAL VOLUME to obtain [A - ]. Calculate K b  (K a x K b = K w ).  Construct an “equilibrium” reaction table, reacting the conjugate base with water.  K b = [OH - ] [BH + ] and obtain [OH - ].  [B]  Calculate the pOH, then the pH.  The equivalence point is ALWAYS >7!

Additions beyond the equivalence point   Construct a “stoichiometry” reaction table.   Determine MOLES of base in excess (not neutralized) and the MOLES of conjugate base.   Divide MOLES by the TOTAL VOLUME,   Because [OH - ] excess >> [OH - ] conj. base, use [OH - ] excess to calculate pOH, then the pH.

Problem   50.0 mL of 0.10 M acetic acid (K a = 1.8 x ) are titrated with 0.10 M NaOH. Calculate the pH after the additions of 0, 10, 25, 40, 50, 60, and 75 mL samples of NaOH.   Then, construct a titration curve and label it properly.