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Simora [160]
2 years ago
5

Calculate the percent dissociation of crotonic acid (C,H,CO,H) in a 0.63 mM aqueous solution of the stuff.

Chemistry
1 answer:
Sauron [17]2 years ago
5 0

Answer:

16%

Explanation:

Crotonic acid : C₃H₅CO₂H

C₃H₅CO₂H  ⇄ C₃H₅CO₂⁻ +  H⁺

   C                      O                O

where : C = C ( 1 - ∝ ) ,   O = C∝

also:  Ka = C∝² / ( 1 - ∝ )  ---- ( 1 )

<em>From Alexa data resource : </em>

Pka = 4.69 ,  [  Ka = 2.04 * 10^-5 = C∝² / ( 1 - ∝ ) ]

back to equation 1

2.04 * 10^-5 = [ ( 0.63 * 10^-3 ) * ∝² / ( 1 - ∝ ) ]  ----- ( 2 )

∴ ∝² / ( 1 - ∝ ) = 3.24 * 10⁻²

Resolving equation above

∝ = 0.1645 =  16.45%

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Calculate the ph of a 0.021 m nacn solution. [ka(hcn) = 4.9  10–10]
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<span>Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN</span>⁻(aq) + Na⁺<span>(aq).
Chemical reaction 2: CN</span>⁻ + H₂O(l) ⇄ HCN(aq) + OH⁻<span>(aq).
c(NaCN) = c(CN</span>⁻<span>) = 0.021 M.
Ka(HCN) =  4.9·10</span>⁻¹⁰<span>.
Kb(CN</span>⁻) = 10⁻¹⁴ ÷ 4.9·10⁻¹⁰ = 2.04·10⁻⁵<span>.
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[HCN] · [OH</span>⁻<span>] = x.
[CN</span>⁻<span>] = 0.021 M - x..
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