Calculate the ph of a 0.200 m nach3co2 solution. ka for acetic acid, ch3co2h, is 1.8 × 10-5.

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.

    NaCH3CO2(aq) + H2O(l) > CH3CO2H(aq) + NaOH(aq)

I     0.2                                               0                          0

C   -x                                                  +x                         +x

E   0.2 - x                                            x                           x

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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a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.

3 1/4 %

explanation: