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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practical process described pest part of scientific investigation.

because practical knowledge is powerful then therotical knowledge.

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.