What is the freezing point of bromine
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Answer:
You will need 9,0 g of acetic acid
Explanation:
The equilibrium acetate-acetic acid is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,76
Using Henderson-Hasselbalch you will obtain:
pH = pka + log₁₀
Where HA is acetic acid and A⁻ is acetate ion
4,90 = 4,76 + log₁₀
1,38 = <em>(1)</em>
As acetate concentration is 0,300M:
0,300M = [HA] + [A⁻] <em>(2)</em>
Replacing (2) in (1):
[HA] = 0,126 M
And:
[A⁻] = 0,174 M
As you need to produce 500 mL:
0,5 L × 0,126 M = 0,063 moles of acetic acid
0,5 L × 0,174 M = 0,087 moles of acetate
To produce moles of acetate from acetic acid:
CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O
Thus, moles of acetate are equivalents to moles of NaOH and all acetates comes from acetic acid, thus:
0,087 moles of acetate + 0,063 moles of acetic acid ≡ 0,15 moles of acetic acid × = <em>9,0 g of acetic acid</em>
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I hope it helps!