pH of Weak Acid Solutions. Example: Calculate the pH of a 0.10 M solution of
nitrous acid HNO. 2 ... equilibrium problem. +. 2. 2. HNO ( ) H ( ) NO ( ) ....
dissociated increases even though the concentrations of all three components
decrease. ( ).
Acids and Bases Chapter 7 Zumdahl 6th Ed. The nature of Acids (Section 7.5) pH of an Acid. (Do the strong and weak cases together). The fraction of dissociation A weak acid and a weak acid in water Solve: for pH and fraction dissociation. Looking ahead: Several weak acids in water.
Strong vs Weak Acids K large
A + − HA ( aq ) + H 2O ( aq ) H aq + A ( ) ( aq ) K A small
Acid
Base
Acid
Base
Strong Acid Most Dissociates
H + A− X2 K A = QA = = HA CA − X [ ]
Each acid has a conjugate base (partner) Each base has a conjugate acid (partner). A strong acid: Mostly Dissociated Weak acid: Little Dissociated
(
pH = − log10 H +
)
and
Weak Acid Little Dissociates
pK A = − log10 ( K A )
Auto-ionization of Water
Equivalent forms
{
2H 2O ( aq ) H 3O + ( aq ) + OH − ( aq ) H 2O ( aq ) H + ( aq ) + OH − ( aq )
Water at equilibrium on its own will have:
QW =
aH + aOH − aH 2 O
= H + ( aq ) OH − ( aq ) = KW = 1.0 ⋅10−14
Pure Water : H + ( aq ) = OH − ( aq ) ∴
pKW = 14
H + ( aq ) = KW
But this means that water must always satisfy the equiliribum when you throw in any other acids or bases into water: Always : H + OH − = KW
Auto-ionization of Water
The reaction is endothermic (because on net a bond is broken), therefore the equilibrium moves to the right as the temperature is increased, more acid is generated. The KW itself increases with temperature. (LeC’s Principle). T − 298 ) ( 1 pKW (T ) = 14 − 9.5* pure water: pH = pKW T 2 At T=50C pH=6.6. It is still neutral water and pH=pOH, but KW increased with the temperature. Therefore it is not acidic. http://www.chemguide.co.uk/physical/acidbaseeqia/kw.html#top What fraction of water molecules ionize? Water is 55 molar and pH=7. + H f = = 18 ⋅10−10 = 2 ppb [ H 2O ]
pH and pOH Scales KW = 1.0 ⋅10−14 = H + OH −
{
}
log10 ( KW ) = log10 (1.0 ⋅10−14 ) = log10 H + OH − pKW = 14 = pH + pOH ⇐ Always pOH = 14 − pH
⇐ Always
For pure water the proton and hydroxide concentrations must be equal; charge balance, charge neutral.
H + > 10−7 Base H + < 10−7 Acid
pH < 7
pOH > 7
pH > 7
pOH < 7
Is an acid hydroxide free? What is the pOH of 10M Hcl? What is the pOH of 1e-8M HCl?
Follow the pH via a chemical Reaction Cr2O7 2− ( aq ) + H 2O ( aq ) 2 H + ( aq ) + 2CrO4 2− ( aq ) orange
yellow 2
H CrO4 KA = Cr2O7 2− +
2−
2
The color is sensitive to the pH and can be used to give us an idea of the KA of the reaction. The Universal pH indicator is a set of dyes that change color over a very wide range of pH
The pH of an Acid (Approximate Treatment) What is the pH of an acid with concentration CA and equilibrium constant KA? NICE Table: HA H + + A− Species HA Initial C A = [ HA]0
H+ 0
A− 0
Change New
X X
X X
−X CA − X
What is the range on X? 0 < X < C A At Equilibrium: Acid Conservation:
H + A− X2 K A = QA = = CA − X [ HA]
A− + [ HA] = X + ( C A − X ) = C A = [ HA]o
The pH of an Acid (Approximate Treatment) + − H = A The answer is X because: = X The range on X: 0 < X < CA Defines two limits: A) X~0, B) X~CA Solve the Quadratic X2 KA = CA − X
⇒
X=
K A2 + 4 K AC A − K A 2
A) Weak Acid K A < CA K A = CA X =
X = 0.62C A
K AC A
Dissociation: Little
~half K
Large
B ) Strong Acid K A > CA X = CA
Almost All
+ − HA H + A K Small
pH of Weak Acid Solutions Example: Calculate the pH of a 0.10 M solution of nitrous acid HNO2 (Ka = 4.0 x 10-4). + −
HNO 2 (aq) H (aq) + NO 2 (aq)
This is a weak acid; therefore we will treat this just like a standard equilibrium problem. + −
H 3O NO2 1 M 1 M H + NO2− = K A = 4 ⋅10−4 = QC = [ HNO2 ] [ HNO2 ] 1M
[ HNO2 ] = 0.1 − x 2
x H = NO = x Final Expression 4 ⋅10 = .1 − x +
− 2
−4
pH of Weak Acid Solutions Example: Calculate the pH of a CA=0.10 M solution of nitrous acid HNO2 (Ka = 4.0 x 10-4). HNO 2 (aq ) H + (aq ) + NO 2− (aq ) Exact solution of the quadratic 2 x 4 ⋅10−4 = , x = 0.00613, .1 − x
pH = px = 2.21
Compare with the approximation: (x CA
X = 0.62C A
X = CA
Determine [H+], and pH for the following examples. 1) HCl, at 0.02M, (stomach acid) KA=10+6 H + = 0.0200; Quad
pH = − log ( 0.02 ) = 1.70
2) Acetic acid, Vinegar, 0.5M, KA=1.6 10-5? (Why can you put Vinegar on salad but HCL will send you to the hospital?) H + = K AC A = 1.6 ⋅10−5 ⋅ 0.05 = 9.0 ⋅10−4 ; Notice : K A < H + < C A
and
pH = 3.05
1 pH = 3.05 = ( pK A + pC A ) 2
Two cases of pH. • • • • • • • •
3) HCl at 1e-8 M. Using the quadratic solution: pH=8. Why is this wrong? Correct Answer: 4) HF 1mM = 0.001M, KA=7.4 10-4. Quadratic answer ans: (5.6 vs 8.6)e-4 Is this a weak acid? CA~KA. Outside the ‘5%’ rule
[ H + ]Quad = 5.6 ⋅10−4 ; pH = 3.25 −4 H + = K C = 8.6 ⋅ 10 [ ] A A • percent dissociation is 56%.
Effect of Dilution on Percent Dissociation and [H+] of a Weak Acid Increasing the concentration of an acid increases all forms, and so the [H+] goes up. But how does the equilibrium partition? One way to see how concentration effects work is to look at the fraction of the acid that is dissociated vs the fraction that is not. − The two fractions must sum to one. [ HA] A − C A = A + [ HA] or 1 = + From our observation about the CA CA conservation of acid − f A−
A HA] [ = and f HA = CA CA
Astoundingly, The fraction (or percent) dissociation is : • INDEPENDENT of the concentration of the acid!!! • depends ONLY on the hydrogen ion concentration. • Decreases ALWAYS with increasing hydrogen ion concentration. (LeC)
Effect of Dilution on Percent Dissociation and [H+] of a Weak Acid K large
A + − HA ( aq ) H aq + A ( ) ( aq ) K small A
X2 KA = CA − X
− [ HA] A − C A = A + [ HA] or 1 = + CA CA
f A−
A− [ HA] = and f HA = CA CA
LeChatelier tells us that if we remove some of the acid (say the HA component) then products will shift to have fewer of both products. But, lowering the concentration is like increasing the volume (which for a gas means decreasing the pressure). This provides more space for the molecules, so this will shift the equilibrium to the right to generate more individual molecules, this means the fraction dissociated increases even though the concentrations of all three components decrease.
The fraction of Dissociation of an Acid HA ( aq ) H + ( aq ) + A− ( aq )
H + A− A− KA KA = or ≡r= H + [ HA] [ HA]
How can the fraction of dissociation be independent of the concentration? Dissociation fraction depends H + A− A− KA ≡r= KA = or only on the proton + HA HA H [ ] [ ] concentration (or pH). A− A− f A− CA r = = ⇒ f A− = r≡ 1 − f A− 1+ r [ HA] [ HA] CA Consistent with the equilibrium expression, and LeChatelier’s Principle: Increase [H+] KA r f A− = = 1 + r K A + H + and the equilibrium moves to left, and fraction A- goes down.
Dissociation distinguishes weak from strong Acid
pH = pK A Dissociation: Little
f A−
