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Ionic Equilibrum question

2010 · Q116
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Ionic Equilibrum question

2010 · Q116

NEETChemistryIonic EquilibrumMCQ+4 / −1
What is [H+] in mol/L of a solution that is 0.20 M in CH3COONa and 0.10 M in CH3COOH? Ka for CH3COOH = 1.8 ×\times× 10−-−5
  1. A
    3.5 ×\times× 10−-−4
  2. B
    1.1 ×\times× 10−-−5
  3. C
    1.8 ×\times× 10−-−5
  4. D
    9.0 ×\times× 10−-−6
View written solutionFree

Correct answer: D

CH3COOH and CH3COONa constitute to form an acidic buffer.

⇒\Rightarrow⇒ pH = pKa + log[Salt][Acid]{{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}[Acid][Salt]​

pH = –log(1.8 × 10–5) + log (0.20)(0.10){{\left( {0.20} \right)} \over {\left( {0.10} \right)}}(0.10)(0.20)​

= 4.74 + log 2

= 4.74 + 0.3010 = 5.041

Now, pH = – log[H+]

⇒\Rightarrow⇒ 5.041 = – log[H+]

⇒\Rightarrow⇒ [H+] = 10–5.041 = 9.0 × 106 mol L–1

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