Strong base neutralizes weak acid Strong acid neutralizes weak base
Jan 17, 2016
Strong base neutralizes weak acid
Strong acid neutralizes weak base
How do we calculate the pH of a buffer ?
2 ingredient problem
0.10 M CH3COOH + 0.20 M CH3COO- pH = 5.05
Consider mixture of salt NaA and weak acid HA.
HA (aq) H+ (aq) + A- (aq)
NaA (s) Na+ (aq) + A- (aq)Ka =
[H+][A-][HA]
[H+] =Ka [HA]
[A-]
-log [H+] = -log Ka - log[HA]
[A-]
-log [H+] = -log Ka + log [A-][HA]
pH = pKa + log [A-][HA]
Remember pX = -logX so pKa = -log Ka
Henderson-Hasselbalch equation
pH = pKa + log[conjugate base]
[acid]
Take log
Assumption: Changes in [A-] & [HA] will be negligible (within 5%) if
Ka < 0.01 & Ka < 0.01 [base] [acid]
use initial conc. of acid and base to calculate pH
What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?
HCOOH (aq) H+ (aq) + HCOO- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.30 0.00
-x +x
0.30 - x
0.52
+x
x 0.52 + x
Common ion effect
0.30 – x 0.30
0.52 + x 0.52
pH = pKa + log [HCOO-][HCOOH]
HCOOH pKa = 3.77
pH = 3.77 + log[0.52][0.30]
= 4.01
16.2
Mixture of weak acid and conjugate base!
A buffer solution is a solution of:
1. A weak acid or a weak base and
2. The salt of the weak acid or weak base
Both must be present!
A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.
16.3
Add strong acid
H+ (aq) + CH3COO- (aq) CH3COOH (aq)
Add strong base
OH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l)
Consider an equal molar mixture of CH3COOH and CH3COONa
Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na2CO3/NaHCO3
(a) KF is a weak acid and F- is its conjugate basebuffer solution
(b) HBr is a strong acidnot a buffer solution
(c) CO32- is a weak base and HCO3
- is it conjugate acidbuffer solution
16.3
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?
NH4+ (aq) H+ (aq) + NH3 (aq)
pH = pKa + log[NH3][NH4
+]pKa = 9.25 pH = 9.25 + log
[0.30][0.36]
= 9.17
Ka = 5.6 x 10-10
=> pOH = 14 - pH = 4.75 [OH-] = 1.8 x 10-5 M
= 9.20
Calculate moles of all components. Remember we have 80.0 mL of the buffer solution @ 0.30 M NH3/0.36 M NH4Cl
and 20.0 mL of 0.050 M NaOH.
NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)
start (moles)
end (moles)
0.029 0.0010 0.024
0.028 0.0 0.025
pH = 9.25 + log[0.25][0.28]
[NH4+] =
0.0280.10
final volume = 80.0 mL + 20.0 mL = 100 mL
[NH3] = 0.0250.10
Moles NH4+
= (0.080 L)(0.36 M) = 0.029 molesMoles NH3 = (0.080 L)(0.30 M) = 0.024 moles
Moles NaOH = (0.020 L)(0.050 M) = 0.0010 moles
Moles/Volume(L)
ΔpH = 0.03
Maintaining the pH of Blood
16.3
How do we build a better buffer?
Add approximately equal quantities of acid and base
Have relatively high concentrations of acid and base
=> the larger the [acid] & [base] the greater the buffer capacity
How do we prepare a buffer at a given pH?
Choose acid/base conjugate pair from table
Check to be sure that they are unreactive in the system used
pKa pH typical rule of thumb
pH + 1 = pKa
Create buffer with pH = 7.50.
pH + 1 = pKa
Look for pKa‘s 6.5 => 8.5
=> Ka‘s 3 x 10-7 => 3 x 10-7
HOCl / OCl Ka = 3.5 x 10-8
H2PO4- / HPO4
-2 Ka2 = 6.2 x 10-8
H2AsO4- / HAsO4
-2 Ka2 = 8 x 10-8
H2CO3 / HCO3- Ka1 = 4.3 x 10-7
Calculate quantities of acid and base.
pH = pKa + log[conjugate base]
[acid]
H2CO3 / HCO3- Ka1 = 4.3 x 10-7
pKa1 = 6.37
pH - pKa = log [HCO3-]
[H2CO3]
7.50 - 6.37 = 1.13
10-1.13 = 13.6 = [HCO3-]
[H2CO3]
Set [H2CO3] = 0.0100 M (NOTE: This is a judgement call.)
then [HCO3-] = 13.6 [H2CO3] = 0.136 M