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4.12 pH Buffer and Buffer Intensity (SJ 146) (Ref: SM p. 54-68 on titration; p.70-73 on buffers) Buffer Solution
- a solution that (in some way) has the ability to maintain a stable composition (conditions)when various components are added or removed
- e.g., pH buffers, metal ion buffers, oxidation-reduction buffers Common pH buffer - a fairly concentrated solution of weak acid and its conjugate base, which do not themselves enter into the reaction
- e.g., a) HAc & NaAc Weak acid HAc ↔ H+ (^) + Ac- Conjugate base NaAc → Na+^ + Ac-^ Ac-^ + H 2 O → HAc + OH- b) NH 4 Cl & NH Weal acid (^3) NH Conjugate base NH^43 Cl + H^ →^2 ONH →^4 + NH+^4 +Cl^ +-^ OH-NH^4 +^ →^ NH^3 +^ H+ c) Na 2 HPO 4 & NaH Weal acid 2 PO 4 NaH 2 PO 4 → Na+ (^) + H 2 PO 4 - (^) H 2 PO 4 - (^) → HPO 4 2- (^) + H+ Conjugate base Na 2 HPO 4 → 2Na+^ + HPO 4 2-^ HPO 4 2-^ + H 2 O → H 2 PO 4 -^ + OH- A solution buffered at a particular pH value will contain:a) an acid that reacts with any strong base added to the solution b) a base that reacts with any strong acid added to the solution Preparation of Buffer The common method of preparing a pH buffer solution is: a)b) Add a mixture of a weak acid and its conjugate baseFor maximum buffer capacity, the pKa of the acid should be very close to the desired pH of the buffer solution.
Example 1 (Butler, p. 145-146) A buffer contains 10 -2^ M HAc (CH 3 COOH) and 10 -2^ M NaAc (CH 3 COONa). (1)(2) Calculate the pH of the buffer.Calculate the pH of the buffer solution when 1 x10-3 (^) mole of HCl is added to 1 L of the buffer solution.
(Solution) (1) Calculate the pH of the buffer Equilibria NaAc → Na+^ + Ac- 10 -2^10 -2^10 - HAc ↔ H+^ + Ac- 10 - H 2 O ↔ H+^ + OH-
[ ][ ]
K [ ]
H Ac a =^ HAc = x
- 10 5 (pKa = 4.75) (1)
Kw = [ H +^ ][ OH −^ ]= 10 −^14 (pKw = 14) (2)
Mass Balance: CT,Ac = [HAc] + [Ac-] = (1 x 10-2) + (1 x 10-2^ )= 2 x 10-2^ (3) CT, Na = [Na+] = 10-2^ (4) Charge Balance: [H+] + [Na+] = [OH-] + [Ac-] (5) From Mass Balance, [Ac-] = 2 x 10-2^ - [HAc] (3)’ Substitute (3)’ and (4) into Charge Balance [H+] + (10-2) = [OH-] + (2 x 10-2^ - [HAc]) (5)’ Thus, [H+] + [HAc] = [OH-] + (10-2) (5)” Since the buffer solutions are fairly concentrated, [H+] << [HAc] and [OH-] << 10- Thus, [HAc] = 10-2^ (6) Substituting (6) into (3)’ yields (7)
Substitute (4) and (5) into Charge Balance [H+] + (10-2) = [Ac-] + [OH-] + ( 10 -3)
- With the assumption: [OH-] ~ 0 [H+] + (10-2) = [Ac-] + (10-3) (5)’ From (5)’ [Ac-] = [H+] + 9 x10-3^ (5)’’ From Mass Balance CT,Ac = [HAc] + [Ac-] = 2 x 10-2^ (3) [HAc] = 2 x 10-2^ - [Ac-] = 2 x 10-2^ - ([H+] + 9 x10-3) [HAc] = 0.011 - [H+] (3)’ Substitute (5)” and (3)’ into (1)
[ ] ( [ ] )
[ ]
K
H H
a = H x
0 011^175
Let [H+] = x and solve for x
x x x x
− =^
x^2 +9.0175 x 10-3^ x -1.925 x 10-7^ = 0
x =−^ (^ 9 0175 10 x^ −^ )^ ±^ (^ 9 0125 10 x^ −^ )^ −^ 4 1^ (^ −1925 10 x − )
. 3. 3 2 ( ).^7
x = −^ 9 0175 10.^ x^^ −^32 ± 9 060 10.^ x^ −^3 =2 125 10. x −^5
x = [H+] = 2.125 x 10- pH = - log (2.125 x 10-5) = 4.67 pH = 4.
ii) With two assumptions: The buffer solution is fairly concentrated, although a)b) the concentration of AcHAc increased by 10%-^ decreased by 10%
Charge Balance: [H+] + [Na+] = [Ac-] + [Cl-] + [OH-] (5) Substitute (4) and (5) into Charge Balance [H+] + (10-2) = [Ac-] + (10-3) - [OH-] (5)’ Assumptions H+] << 10-2^ and [OH-] << [Ac-] + (10-3) Thus, [Ac-] = 10 -2^ - 10-3^ = 0.009 (6) From Mass Balance, CT,Ac = [HAc] + [Ac-] = 2 x 10-2^ (3) Substituting (6) into (3) yields [HAc] + 0.009 = 2 x 10- [HAc] = (2 x 10-2^ ) - (9 x 10-3^ ) = 1.1 x 10-2^ (7) Substituting (6) and (7) into (1) yields
[ ][ ] [ ]
Ka = H^ x x = x
9 10 11 10 175 10
3
.^2.^5 (6)
[^ H ] (^ ( )(^ ) )
x x x x
+ = 175 10−^ −11 10− = −
5 2 3 5
[H+] = 2.14 x 10-5^ pH = 4.
{ [ ] [ ]} [ ] [ ] K (^) a = H^^ + HA^ A −^ = H^ C (^) HAC^ NaA + OH +^ H −− HOH
- − [ (^) [ ][ (^) ] ] − + [ ] (11)
Rearrange for [H+]
[ ] [ ] [ ] [ ]
[ H ] K aCC HA OHOH^ HH NaA
This is a general expression for pH of the buffer solution. For most buffer solutions, CHA >> ( [OH-] - [H+] ) and CNaA >> ( [OH-] + [H+] ) Eq. (12) reduces to
[ H^ +^ ]= K (^) a (^) CCNaAHA (13)
This is Henderson’s eqn. Taking logarithm of Eq. (13) yields
log[ H^ +^ ] = log K (^) a +log CCNaAHA or − log[ H +] = − log K (^) a −log CCNaAHA
pH = pK (^) a − log CCNaAHA
Thus, pH = pK (^) a + log CCNaAHA a log A HA pH pK C C
[ ] pH pK (^) [ ] salt = (^) a + log (^) acid log a conjugate base weak acid pH pK C = + C
This equation is commonly used for the preparation of buffer solution at a given pH.
Example : A buffer contains 10 -2^ M HAc and 10 -2^ M NaAc.
- Calculate the pH of the buffer solution
- Calculate the pH of the solution whensolution. 1.0 x 10-3^ moles of HCl is added to 1L of the buffer
(Solution) Equilibria NaAc → Na+^ + Ac- 10 -2^10 -2^10 - HAc 10 -2 ↔ H+^ + Ac-^ pKa = 4.
CNaA = 0.01 M and CHA = 0.01 M Using the Henderson’s eqn , pH = pK (^) a + log CCAHA^ − = 4 75. + log 0 010 01.. =4 75. (^) pH = 4.
Equilibria after the addition of 0.001 M HCl, NaAc → Na+^ + Ac- 0.01 0.01 0.01 – 0. HAc ↔ H+^ + Ac- 0.01 + 0.
pH pK ( )
C
a C
A
= + HA = +
log −. log.^.
pH = 4 75. + log 0 009 0.. 011 =4 66. pH = 4.
- Calculate the pH of the buffer solution.
pH = pK (^) a + log^ CCconjugate baseweak acid a log A HA pH pK C C
42 2 4
log 6.7 log 0.2 6.7 0.3 7.
a^ HPO H PO pH pK C C
− = + (^) −= + = + = (^) pH = 7
- Calculate the pH of the solution when 0.05 mole of NaOH is added to 1L of the buffer solution. NaOH → Na+^ + OH- 0.05 0.05 0. H 2 PO 4 -^ + OH-^ → HPO 4 2--^ + H 2 O
K 2 HPO 4 → 2K+^ + HPO 4 2- 0.2 0.2 0.2 + 0. KH 2 PO 4 → K+^ + H 2 PO 4 - 0.1 0.1 0.1 – 0. pH = pK (^) a + log^ CCconjugate baseweak acid a log A HA pH pK C C
42 2 4 log 6.7 log 0.2^ 0.05 6.7 0.7 7. 0.1 0. a^ HPO H PO pH pK C C
− − = + = + + = + = − pH = 7.
Thus, pH of the buffer solution would rise only from 7.0 to 7.4. If the same quantity of base had been added to distilled water, the pH would have risen above 12. NaOH 0.05 → Na0.05+^ + (^) 0.05OH-
pOH = -log (0.05) = 1. pH = 14- 1.3 = 12.
