Ionic equilibrium

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Question

When 0.2 M solution of acetic acid is neutralised with 0.2 M NaOH in 500 mL of water, the pH of the resulting solution will be $[p K_{a} of acetic acid = 4.74]$

Moderate

Solution

${CH}_{3} COOH + NaOH ⟶ {CH}_{3} COONa + H_{2} O$
$\Rightarrow$ At equivalence point, a salt of weak acid, strong base is formed.
$\Rightarrow pH = 7 + \frac{1}{2} (p K_{a} + \log C)$
$Here C = concentration of salt = \frac{0.2 \times 500}{500 + 500} = 0.1 M$
[Check that 500 mL of CH₃COOH is also required]
$\Rightarrow pH = 7 + \frac{1}{2} (4.74 + \log 0.1) = 8.87$

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Ionic equilibrium

Unlock the full solution & master the concept.

When 0.2 M solution of acetic acid is neutralised with 0.2 M NaOH in 500 mL of water, the pH of the resulting solution will bepKa of acetic acid = 4.74]

CH3COOH+NaOH⟶CH3COONa+H2O⇒At equivalence point, a salt of weak acid, strong base is formed.⇒pH=7+12pKa+log⁡C Here C= concentration of salt =0.2×500500+500=0.1M[Check that 500 mL of CH3COOH is also required]⇒pH=7+12(4.74+log⁡0.1)=8.87

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When 0.2 M solution of acetic acid is neutralised with 0.2 M NaOH in 500 mL of water, the pH of the resulting solution will be $[p K_{a} of acetic acid = 4.74]$