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