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Ionic Equilibrium question

2021 · 18 Mar · Shift 1 · Q16
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  5. /2021 · 18 Mar · Shift 1 · Q16

Ionic Equilibrium question

2021 · 18 Mar · Shift 1 · Q16

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
In order to prepare a buffer solution of pH 5.74, sodium acetate is added to acetic acid. If the concentration of acetic acid in the buffer is 1.0 M, the concentration of sodium acetate in the buffer is ‾\underline{\hspace{2cm}}​ M. (Round off to the Nearest Integer). [Given : pKa (acetic acid) = 4.74]
Numerical answer
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Correct answer: 10

  1. Use Henderson–Hasselbalch equation

For an acidic buffer,

pH=pKa+log⁡[salt][acid]\text{pH} = \text{p}K_a + \log\frac{[\text{salt}]}{[\text{acid}]}pH=pKa​+log[acid][salt]​

Here,

  • pH=5.74\text{pH} = 5.74pH=5.74
  • pKa=4.74\text{p}K_a = 4.74pKa​=4.74
  • [acid]=[CH3COOH]=1.0 M[\text{acid}] = [\text{CH}_3\text{COOH}] = 1.0\,\text{M}[acid]=[CH3​COOH]=1.0M
  • [salt]=[CH3COONa]=?[\text{salt}] = [\text{CH}_3\text{COONa}] = ?[salt]=[CH3​COONa]=?
  1. Substitute the values

5.74=4.74+log⁡[salt]1.05.74 = 4.74 + \log\frac{[\text{salt}]}{1.0}5.74=4.74+log1.0[salt]​

5.74−4.74=log⁡[salt]5.74 - 4.74 = \log[\text{salt}]5.74−4.74=log[salt]

1.00=log⁡[salt]1.00 = \log[\text{salt}]1.00=log[salt]

  1. Convert from logarithm

[salt]=101.00=10 M[\text{salt}] = 10^{1.00} = 10\,\text{M}[salt]=101.00=10M

  1. Nearest integer

10 M10\,\text{M}10M

So, the concentration of sodium acetate required is 10 M10\,\text{M}10M.

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