# Buffers

Buffers

COMPUTATIONS

Table A. pH Way of measuring using pH meter

Worked out pH

Option 1 – HoAc

zero. 10 M CH3COOH

CH3COOH + H2O ⇌ CH3COO- & H3O+

we 0. 12 ø ø

c -x +x +x

elizabeth 0. 15 – by x back button

Ka sama dengan H3O+[CH3COO-]CH3COOH sama dengan x20. twelve – by = 1 ) 8 times 10-5 back button = 1 . 33 x 10-3 M

pH sama dengan -log [1. thirty-three x 10-3]

pH sama dengan 2 . 88

Solution two – HoAc – OAc

na + nb sama dengan nbuffer

= zero. 05 M (0. 12 L)

em + nb = 0. 005 mol

pH sama dengan 5

ph level = pKa + journal nbna

five = 4. 74 + log nbna

log nbna = 0. 026 nbna = 1 . 82

mhh + 1 . 82na sama dengan 0. 005 mol

bist du = 1 ) 77 by 10-3 mol

nb = 3. 23 x 10-3 mol

Se till att du ar = 1 . 77 by 10-3 mol * (1/0. 2 mol) = eight. 85 back button 10-3 D

Vb sama dengan 3. 3 x 10-3 mol * (1/0. a couple of mol) sama dengan 0. 0162 L

ph level = four. 74 + log (0. 2M/8. eighty five x 10-3 L)(0. 2M/0. 0162 T

pH sama dengan 5

Solution 3 – NH3

0. 10 M NH3

NH3 + INGESTING WATER ⇌ NH4+ + OH-

i 0. 10 ø ø

c -x +x +x

e zero. 10 – x by x

Kilobytes = x20. 10 – x sama dengan 1 . almost 8 x 10-5

x = 1 . thirty-three x 10-5 M

pOH = -log [1. 33 by 10-5]

pOH = 2 . 88

pH sama dengan 11. 12

Solution four – NH3 – NH4+

na & nb = nbuffer

= 0. 05 M (0. 10 L)

bist du + nb = 0. 005 mol

pH sama dengan 9, pOH = a few

pOH = pKa & log nanb

5 = 4. 74 + log nanb

sign nanb sama dengan 0. 026 nanb sama dengan 1 . 82

nb & 1 . 82nb = 0. 005 mol

nb = 1 . 77 x 10-3 mol

mhh = a few. 23 x 10-3 mol

Va = 3. twenty-three x 10-3 mol 2. (1/0. two mol) sama dengan 0. 0162 L

Vb = 1 . 77 times 10-3 mol * (1/0. 2 mol) = eight. 85 times 10-3 D

pOH = 4. 74 + journal (0. 2M/8. 85 by 10-3 L)(0. 2M/0. 0162 L

pOH = your five

pH = 9

Solution 5 -- NaH2PO4

0. 10 M NaH2PO4

ph level = 12(pKa1 + pKa2)

pH sama dengan 4. 67

Percent Mistake

% error= calculated-measuredcalculated ×100

Sample(solution 1)

% error= 2 . 88-2. 842. 88 ×100 =1. 39%

Desk B.

Solution you – HOAc

a. addition of acidity

15 cubic centimeters 0. 10 M CH3COOH + 0. 1 milliliters 1 M HCl

CH3COOH & H2O ⇌ CH3COO- + H3O+

my spouse and i 0. 10M(15 ml)15. 1 ml ø 1M(0. one particular ml)15. 1 ml c -x +x +x

e zero. 10M(15 ml)15. 1 ml-x x 1M(0. 1 ml)15. 1 ml+x

Ka = H3O+[CH3COO-]CH3COOH sama dengan x+(1M0. one particular ml15. one particular ml+x)0. 10M(15 ml)15. you ml-x sama dengan 1 . almost 8 x 10-5 x sama dengan 2 . fifty nine x 10-4 M

ph level = -log [1M(0. 1 ml)15. 1 ml+2. 59 times 10-4] pH = 2 . 18

b. addition of foundation

15 cubic centimeters 0. 15 M CH3COOH + 0. 1 cubic centimeters 1 M NaOH

CH3COOH + OH- ⇌ CH3COO- + H2O we 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. one particular ml)15. you ml +1M(0. 1 ml)15. 1 cubic centimeters

e 0. 10M(15 ml)15. 1 ml-1M(0. one particular ml)15. you ml 0 1M(0. one particular ml)15. one particular ml Applying Henderson-Hasselbach Formula:

pH=pKa+logBaseAcid

pH=4. 74+log1M(0. one particular ml)15. one particular ml 0. 10M(15 ml)15. 1 ml- 1M(0. one particular ml)15. one particular ml

pH sama dengan 3. 59

Solution 2 – HOAC-OAC- buffer

a. addition of acid

pH=pKa+lognbase- nsanacid+ no-strings-attached

pH=4. 74+log3. 23×10-3- 1×10-41. 77×10-3+ 1×10-4

pH sama dengan 4. ninety six

b. addition of foundation

pH=pKa+lognbase+ nsbnacid- nsb

pH=4. 74+log3. 23×10-3+ 1×10-41. 77×10-3- 1×10-4

pH = 5. 04

Option 3- NH3

a. addition of acid solution

15 milliliters 0. 15 M NH3 + 0. 1 milliliters 1 Meters HCl

NH3 + H+ ⇌ NH4+ i 0. 10M(15 ml)15. 1 cubic centimeters 1M(0. one particular ml)15. you ml ø

c - 1M(0. 1 ml)15. 1 milliliters - 1M(0. 1 ml)15. 1 cubic centimeters +1M(0. one particular ml)15. 1 ml

e 0. 10M(15 ml)15. 1 ml-1M(0. 1 ml)15. 1 cubic centimeters 0 1M(0. 1 ml)15. 1 milliliters

Applying Henderson-Hasselbach Formula:

pOH=pKb+logAcidBase

pOH=4. 74+log1M(0. you ml)15. 1 ml 0. 10M(15 ml)15. 1 ml- 1M(0. one particular ml)15. you ml...

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