# Buffers

**Topics:**PH, Acid dissociation constant, Buffer solution

**Pages:**7 (1093 words)

**Published:**February 7, 2013

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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