Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
Answer:
i know the problem
2c2h6(g)+702(g)->4co2(G)+6g20(1) thats double replacement i dont know the first part
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Hydrogen ion reacts with zinc to produce
gas
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