Class Notes (808,754)
Chemistry (260)
ENCH 213 (25)
Lecture

# 21monoproticacids.pdf

5 Pages
78 Views

School
Queen's University
Department
Chemistry
Course
ENCH 213
Professor
Diane Beauchemin
Semester
Fall

Description
Monoprotic acid-base equilibria • Strong acids and bases • Weak acids and bases ◦ Fraction of dissociation of an acid • Buffers ◦ Henderson-Hasselbalch equation ◦ Buffer capacity If one measures the equilibrium constant for each of the following solutions, which relationship is correct? A- mix CH CH 2H CH2COOH2with2CH CH CH NH (weak 2/ we2k) 2 2 B - mix HNO with3CsOH (strong w/ strong) C - mix KOH with HCOOH (strong w/ weak) a) K for A> K for B > K for C b) K for C > K for A> K for B c) K for B > K for C > K forA pH of pure water + -7 • pH = -log [H O3] = -log 10 = 7 • is the pH of distilled water = 7? H 2 • CO 2 (g)H O2⇔ H CO ⇔2H O 3 HCO 3 + 3- carbonic acid bicarbonate • pH < 7 unless distilled under inert atmosphere (N ) or b2iled and then put under inert atmosphere pH of a strong acid or base solution Strong acid • [acid] = [H 3 ] since complete dissociation • pH=-log a = -log[acid]γ H3O+ + H3O+ • pH ≈ -log[H O 3= -log[acid] Strong base • [OH]=[base] since complete dissociation - - • pOH=-log aOH = -log([base]γOH) • pOH ≈ -log[OH]= -log[base] o • pH = 14.00-pOH (at 25 C) • pH of 1.0 × 10-8 M KOH? -8 ◦ pOH ≈ -log(1x10 ) = 8.00 ◦ pH = 14 – pOH = 6.00 ! Watch out for the acid or base concentration • If ≥ 10 M: ◦ pH determined by added [H O ] or [OH3] + - -8 • If ≤ 10 M: ◦ pH = 7.00 ◦ Autoprotolysis of water predominates • If =10 -10 M:-8 ◦ systematic equilibrium calculation ◦ similar contribution as from water pH of 1.0×10-8 M KOH • Pertinent reactions ◦ 2H O ⇔2H O + OH3 + - Kw=1.0×10 -14 • Charge balance ◦ [K ]+ [H O ] 3 [OH] - • Mass balance ◦ [K ]= 1.0×10 M -8 + - -14 • Equilibrium constant expression: Kw=[H O ][OH]=1.3×10 • Count: 3 unknowns ([K ],[H O ],[OH3) and 3 equations • Solve pH of 1.0×10-8 M KOH • Set x = [H O3], as we seek the pH - + -14 • [OH] = Kw/[H O ]= 3.0×10 /x • [K ]= 1.0×10 M -8 • Insert into charge balance: • 1.0×10 + x = 1.0×10 -14/x 2 -8 -14 • x +(1.0×10 )x -(1.0×10 )=0 quadratic equation 8 • pH=-log (9.6 ×10- M) = 7.02 -7 + - Water almost never produces 10 M H O & OH 3 Example: in 10 M HBr -4 • pH = -log (10 ) = 4.00 • pOH = 14.00 - 4.00 = 10.00 -10 • [OH-] =10 M only comes from water • Water produces 10 -10M OH and 10 -10M H O 3 + Le Châtelier’s principle acts on water too → adding acid or base shifts the autoprotolysis equilibrium to the left (more water) pH of weak acid solution • pertinent reactions and equilibrium constants + - + - ◦ HA+ H O ⇔ H 2 +A 3 K a[H O 3[A]/[HA] ◦ 2 H O ⇔ H O + OH + - K =[H O ][OH] - 2 3 w 3 • charge balance: [H O ] 3 [A] + [OH] - - • Mass balance: C = [Aa + [HA] • If C a 10 M -6 - - ◦ [OH ] << [A] + - ◦ [H O ]3≈ [A] = x and 2 ◦ i.e. x + K x – a C = 0a qaadratic equation ◦ ◦ pH = -log x • **retain all digits in the calculator, and express the pH to the 0.01 decimal place Example: pH of 0.100 M (CH ) NHCl ? 3 3 • (CH ) NHCl is a salt that completely dissociates into CH ) NH and Cl + - 3 3 3 3 • (CH )3 3 + H O ⇔ (C2 ) N + H3O 3 3 + ca=0 100M 0 0 initially ca– x
More Less

Related notes for ENCH 213

OR

Don't have an account?

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.

Get notes from the top students in your class.

Request Course
Submit