Buffers

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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...
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