What is [H3O +] in a solution of 0.25 M CH3CO2H and 0.030 M NaCH3CO2? ?

            CH₃CO₂H(aq) + H₂O(ℓ) ⇌ CH₃CO₂⁻(aq) + H₃O⁺(aq)    Ka = 1.8 × 10⁻⁵
Initial:      0.25 M                             0.030 M            0 M
Change:    -y M                                  +y M            +y M
Eqm:     (0.25 - y) M                      (0.030 +y) M       y M
               ≈ 0.25 M                           ≈ 0.030 M

At equilibrium:
Ka = [CH₃CO₂⁻] [H₃O⁺] / [CH₃CO₂H]
1.8 × 10⁻⁵ = 0.030y / 0.25
y = 0.25 × 1.8 × 10⁻⁵ / 0.030 = 1.5 × 10⁻⁴
[H₃O⁺] = 1.5 × 10⁻⁴

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OR:

Henderson-Hesselbalch equation:
pH = pKa + log([CH₃CO₂⁻]/[CH₃CO₂H])
pH = -log(1.8 × 10⁻⁵) + log(0.030/0.25) = 3.824
[H₃O⁺] = 10⁻³·⁸²⁴ M = 1.5 × 10⁻⁴ M

Given, the buffer representation,

CH3CO2H(aq) + H2O(l) \rightleftharpoons H3O+(aq) + CH3CO2-(aq)

Also given,

[CH3CO2H] = 0.25 M

[NaCH3CO2-] Or [CH3CO2-]= 0.030 M

Ka= 1.8 x 10-5

We know, the Henderson-Hasselbalch equation to calculate the pH of a buffer solution,

pH = pKa + log[ C.Base / Acid]

pH = -log[1.8x 10-5] + log[ CH3CO2- /CH3CO2H]

pH = -log[1.8x 10-5] + log[ 0.030/0.25]

pH = 4.74 + (-0.92)

pH = 3.82

Now, We know,

pH = -log[H3O+]

Rearranging the formula,

[H3O+] = 10-pH

[H3O+] = 10-3.82

[H3O+] = 1.5 x 10-4 M [ 2S.F]

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