Answer:
12 g of choloracetic acid
Explanation:
The buffer equilibrium is:
HCH₂ClCO₂ ⇄ CH₂ClCO₂⁻ + H⁺
pka= -log ka =
Ka: 1,3x10⁻³ = [CH₂ClCO₂⁻] [H⁺] / [HCH₂ClCO₂]
By Henderson-Hasselbalch equation:
pH = pka + log₁₀ [A⁻] / [HA]
3,01 = 2,89 + log₁₀ [A⁻] / [HA]
1,318 = [A⁻] / [HA]
As molar concentration of chloroacetic acid (HA) is 0,20M
[A⁻] = 0,26 mol/L
The volume is 500 mL ≡ 0,5 L
0,26mol/L × 0,5 L = 0,13 moles of chloroacetic acid. In grams:
0,13 mol × (94,5g / 1mol) = <em>12 g of choloracetic acid</em>
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I hope it helps!
