Respuesta :
Annswer : The correct answer is : pH = 9.03
To calculating pH of NaCH₃CO₂ ,we need to do following steps :
Step 1 : Find kb for NaCH₃CO₂
Since NaCH₃CO₂ and CH₃CO₂H are conjugates base -acid pair , we can calculate kb for NaCH₃CO₂ from ka of CH₃CO₂H using relation :
Kw = ka x kb ,
Where Kw = ionic product of water = 1 x 10 ⁻¹⁴
ka = dissociation constant of acid = 1.8 x 10⁻⁵
kb = dissociation constant of base
Plugging value of kw and ka in formula :
1 x 10 ⁻¹⁴ = 1.8 x 10⁻⁵ x kb
Dividing both side by 1.8 x 10⁻⁵
1 x 10⁻¹⁴ / 1.8 x 10⁻⁵ = 1.8 x 10⁻⁵ / 1.8 x 10⁻⁵ x kb
kb = 5.6 x 10 ⁻¹⁰
Step 2 : Setting ICE ( Initial , change and equilibrium concentration ) Table.
NaCH₃CO₂is strong base , so its ion CH₃CO₂⁻ undergoes hydrolysis and gives OH⁻ as follows:
CH₃CO₂⁻ + H₂O → CH₃CO₂H + OH⁻ .
Initial (M) 0.200 - 0 0
Change(M) -x - +x +x
Equilibrium(M) 0.200 -x x x
Step 3: Writing kb for reaction
kb = [CH₃CO₂H] [OH⁻] / [CH₃CO₂⁻ ]
Plugging equilibrium concentrations in above equation
5.6 x 10 ⁻¹⁰ = [ x] [x] / [0.200 - x ]
Neglecting x from 0.200 -x
5.6 x 10 ⁻¹⁰ = [ x²] / [0.200 ]
Multiplying both side by 0.200
5.6 x 10 ⁻¹⁰ x 0.200 = [ x²] [0.200 ] / [0.200 ]
x² = 1.12 x 10⁻10
x = 1.06 x 10⁻⁵
[CH₃CO₂H] = [OH⁻] = x = 1.06 x 10⁻⁵
Step 4 : Calculating pOH
pOH = - log [OH-]
pOH = -log [ 1.06 x 10⁻⁵ ]
pOH = 4.97
Step 5: Calculating pH
Relation between pH and pOH : pH + pOH = 14
pH = 14 - pOH = 14 - 4.97
pH = 9.03
The pH of the solution obtained is 9.04.
- The equation of the reaction is;
NaCH3CO2(aq) + H2O(l) ------> CH3CO2H(aq) + NaOH(aq)
I 0.2 0 0
C -x +x +x
E 0.2 - x x x
- Since; kw = ka. kb
kb = kw/ka = 1 × 10^-14/1.8 × 10-5
kb = 5.6 × 10^-10
5.6 × 10^-10 = x^2/0.2 - x
5.6 × 10^-10(0.2 - x ) = x^2
1.12 × 10^-10 - 5.6 × 10^-10x = x^2
x^2 + 5.6 × 10^-10x - 1.12 × 10^-10 = 0
x = 0.000011 M
Now;
pOH = - log(0.000011 M)
pOH = 4.96
pH = 14 - 4.96
pH = 9.04
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