Buffers

Buffers

CALCULATIONS

Table A. pH Measurement using pH meter

Calculated pH

Solution 1 – HoAc 0.10 M CH3COOH CH3COOH + H2O ⇌ CH3COO- + H3O+ i 0.10 ø ø c -x +x +x e 0.10 – x x x

Ka = H3O+[CH3COO-]CH3COOH = x20.10 – x = 1.8 x 10-5 x = 1.33 x 10-3 M pH = -log [1.33 x 10-3] pH = 2.88

Solution 2 – HoAc – OAc na + nb = nbuffer = 0.05 M (0.10 L) na + nb = 0.005 mol pH = 5 pH = pKa + log nbna 5 = 4.74 + log nbna

log nbna = 0.026 nbna = 1.82 na + 1.82na = 0.005 mol na = 1.77 x 10-3 mol nb = 3.23 x 10-3 mol

Va = 1.77 x 10-3 mol * (1/0.2 mol) = 8.85 x 10-3 L
Vb = 3.23 x 10-3 mol * (1/0.2 mol) = 0.0162 L

pH = 4.74 + log (0.2M/8.85 x 10-3 L)(0.2M/0.0162 L pH = 5

Solution 3 – NH3 0.10 M NH3 NH3 + H2O ⇌ NH4+ + OH- i 0.10 ø ø c -x +x +x e 0.10 – x x x

Kb = x20.10 – x = 1.8 x 10-5 x = 1.33 x 10-5 M pOH = -log [1.33 x 10-5] pOH = 2.88 pH = 11.12

Solution 4 – NH3 – NH4+ na + nb = nbuffer = 0.05 M (0.10 L) na + nb = 0.005 mol pH = 9, pOH = 5 pOH = pKa + log nanb 5 = 4.74 + log nanb

log nanb = 0.026 nanb = 1.82 nb + 1.82nb = 0.005 mol nb = 1.77 x 10-3 mol na = 3.23 x 10-3 mol

Va = 3.23 x 10-3 mol * (1/0.2 mol) = 0.0162 L
Vb = 1.77 x 10-3 mol * (1/0.2 mol) = 8.85 x 10-3 L

pOH = 4.74 + log (0.2M/8.85 x 10-3 L)(0.2M/0.0162 L pOH = 5 pH = 9

Solution 5 - NaH2PO4 0.10 M NaH2PO4 pH = 12(pKa1 + pKa2) pH = 4.67
Percent Error
% error= calculated-measuredcalculated ×100
Sample(solution 1)
% error= 2.88-2.842.88 ×100 =1.39%

Table B.

Solution 1 – HOAc

a. addition of acid

15 ml 0.10 M CH3COOH + 0.1 ml 1 M HCl CH3COOH + H2O ⇌ CH3COO- + H3O+ i 0.10M(15 ml)15.1 ml ø 1M(0.1 ml)15.1 ml c -x +x +x e 0.10M(15 ml)15.1 ml-x x 1M(0.1 ml)15.1 ml+x

Ka = H3O+[CH3COO-]CH3COOH = x+(1M0.1 ml15.1 ml+x)0.10M(15 ml)15.1 ml-x = 1.8 x 10-5 x = 2.59 x 10-4 M pH = -log [1M(0.1 ml)15.1 ml+2.59 x 10-4] pH = 2.16

b. addition of base

15 ml 0.10 M CH3COOH + 0.1 ml 1 M NaOH CH3COOH + OH- ⇌ CH3COO- + H2O i 0.10M(15 ml)15.1 ml 1M(0.1 ml)15.1 ml ø

c - 1M(0.1 ml)15.1 ml - 1M(0.1 ml)15.1 ml +1M(0.1 ml)15.1 ml

e 0.10M(15 ml)15.1 ml-1M(0.1 ml)15.1 ml 0 1M(0.1 ml)15.1 ml
Using Henderson-Hasselbach Equation:

pH=pKa+logBaseAcid

pH=4.74+log1M(0.1 ml)15.1 ml 0.10M(15 ml)15.1 ml- 1M(0.1 ml)15.1 ml

pH = 3.59

Solution 2 – HOAC-OAC- buffer

a. addition of acid

pH=pKa+lognbase- nsanacid+ nsa

pH=4.74+log3.23×10-3- 1×10-41.77×10-3+ 1×10-4

pH = 4.96

b. addition of base

pH=pKa+lognbase+ nsbnacid- nsb

pH=4.74+log3.23×10-3+ 1×10-41.77×10-3- 1×10-4

pH = 5.04

Solution 3- NH3

a. addition of acid

15 ml 0.10 M NH3 + 0.1 ml 1 M HCl NH3 + H+ ⇌ NH4+ i 0.10M(15 ml)15.1 ml 1M(0.1 ml)15.1 ml ø

c - 1M(0.1 ml)15.1 ml - 1M(0.1 ml)15.1 ml +1M(0.1 ml)15.1 ml

e 0.10M(15 ml)15.1 ml-1M(0.1 ml)15.1 ml 0 1M(0.1 ml)15.1 ml

Using Henderson-Hasselbach Equation:

pOH=pKb+logAcidBase

pOH=4.74+log1M(0.1 ml)15.1 ml 0.10M(15 ml)15.1 ml- 1M(0.1 ml)15.1 ml

pOH = 3.59 pH = 10.41

b. addition of base

15 ml 0.10 M NH3 + 0.1 ml 1 M NaOH NH3 + H2O ⇌ NH4+ + OH- i 0.10M(15 ml)15.1 ml ø 1M(0.1 ml)15.1 ml c -x +x +x e 0.10M(15 ml)15.1 ml-x x 1M(0.1 ml)15.1 ml+x

Kb = H3O+[CH3COO-]CH3COOH = x+(1M0.1 ml15.1 ml+x)0.10M(15 ml)15.1 ml-x = 1.8 x 10-5 x = 2.59 x 10-4 M pOH = -log [1M(0.1 ml)15.1 ml+2.59 x 10-4] pOH = 2.16 pH = 11.84

Solution 4- NH3-NH4+ buffer

a. addition of acid

pOH=pKb+lognacid+ nsanbase- nsa

pOH=4.74+log3.23×10-3+ 1×10-41.77×10-3- 1×10-4

pOH = 5.04 pH= 8.96

b. addition of base pOH=pKb+lognacid- nsbnbase+ nsb

pOH=4.74+log3.23×10-3- 1×10-41.77×10-3+ 1×10-4

pOH = 4.96 pH = 9.04

ANSWERS TO QUESTIONS

1. pH of 0.10 M chloroacetic acid and 0.15 M sodium chloroacetate

pKa= -logKa pKa= -log1.4×10-3 pKa = 2.85

pH=pKa+logBaseAcid pH=2.85+log0.150.10 pH = 3.03

5. Substance | Ka | pKa | HCOOH | 1.8 x 10-4 | -log(1.8 x 10-4) = 3.74 | HClO | 3.5 x 10-8 | -log(3.5 x 10-8) = 7.46 | H2CO3 | 4.2 x 10-7 | -log(4.2 x 10-7) = 6.38 | NH3 | 5.56 x 10-10 | -log(5.56 x 10-10) = 9.255 | (CH3)3N | 1.35 x 10-10 | -log(1.35 x 10-10) = 9.870 |

The best buffer system to use is the buffer system with HClO because its pKa is the closest to 7.00 with the value of 7.46