Class Notes (1,100,000)
CA (650,000)
Queen's (10,000)
CHEM (300)
CHEM 112 (100)
Lecture 2

CHEM 112 Lecture Notes - Lecture 2: Deprotonation, Quadratic Equation, Hydrogen Bond


Department
Chemistry
Course Code
CHEM 112
Professor
John Carran
Lecture
2

This preview shows half of the first page. to view the full 2 pages of the document.
Lecture 2, Week 23
Polyprotic Acids
-polyprotic aid: compound which can donate more than one proton
-have 2 hydrogens
- The titrations of polyprotic acids with a strong base are similar to that of a monoprotic acid, except
that there are additional stoichiometric points associated with the addition acidic hydrogens.
-pKa values are horizontal slopes on pH vs volume graphs
-in the middle of the buffer zone, pH=pKa
-vertical slopes are stoichiometric point for equilivance points
Diprotic hydroxybenzoic aids
-the hydrogen bonding between carboxylates oxygen and the hydrogen on the hydroxyl group stabalize
this species towards dissociation of the proton.
Sulfuric Acid
-Sulfuric acid is a diprotic acid, with the first Ka value being very high (>> 1), such that one of the two
protons is completely dissociated in aqueous solution and the conentration of H3O+ (aq) produced from
the first deprotonation is equal to the original concentration of H2 SO4 .
H2 SO4 (aq) + H2O(l) HSO4 - (aq) + H3O+ (aq)
-The conjugate base, HSO4 - , is a weak acid (pKa = 1.92), such that the second equilibrium involving
deprotonation of HSO4 - , also contributes to the concentration of H3O+ .
HSO4 - (aq) + H2O(l)  SO4 2- (aq) + H3O+ (aq)
-If we have a 0.010 M solution of H2 SO4 (aq), what is the pH of the solution?
-The initial concentration of H3O+ (and HSO4 - ) is 0.010 mol L-1 from the first deprotonation step, and
we will let x be the concentration of H3O+ (and SO4 2- ) from the second deprotonation step, for which
the equilibrium constant is Ka2 = 1.2 x 10-2 .
1.2 x 10-2 = x (.01+x)/(.01-x)
- We can solve for x using the quadratic equation, and find that x = 4.3 x 10-3 . This means that the final
[H3O]+ = 0.014 mol L-1 , which corresponds to a pH of 1.9.
- Note that if only the first acid dissociation took place, the pH, corresponding to [H3O+ ] = 0.010 mol L-1
, would have been 2.0.
How to Calculate the Concentrations of All Species in a Polyprotic Acid Solution Procedure for a Diprotic
Acid (H2A)
Step 1: From the deprotonation equilibrium of the acid (H2A), determine the concentrations of the
conjugate base HA- and H3O+ as was done for a monoprotic acid.
Step 2: Find the concentration of A2- from the second deprotonation equilibrium (that of HA- ) by
substituting the concentrations of H3O+ and HAfrom step 1 into the expression for Ka2.
Step 3: Find the concentration of OH- by dividing Kw by the concentration of H3O+ .
Procedure for a Triprotic Acid (H3A)
Step 1: From the deprotonation equilibrium of the acid (H3A), determine the concentrations of the
conjugate base H2A - and H3O+ as was done for a monoprotic acid.
Step 2: Find the concentration of HA2- from the second deprotonation equilibrium (that of H2A - ) by
substituting the concentrations of H3O+ and H2A - from step 1 into the expression for Ka2.
find more resources at oneclass.com
find more resources at oneclass.com
You're Reading a Preview

Unlock to view full version