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The acid dissociation constant(Ka) is 0.0095
The reaction for this dissociation of acid is
HSO4 ⇄H+ + SO4 -2
The dissociation constant can be determined from the following expression
[H+] = 10-PH
= 10-1.35
[H+] = 0.0447
[H+] = 0.0447 mol / L
From equation, [H+] = [SO4-2] = 0.0447 mol / L
[SO4-2] = 0.0447 mol / L
From the values of [H+], [SO4-2] and [HSO4] Ka can be calculated as follows,
Ka = 0.0447 * 0.0447 / 0.200
= 0.0019 / 0.200
= 0.0095
Hence the value of the acid dissociation constant (Ka) for the given reaction is 0.0095
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Answer:
(a) sp³ sp³
H₃<u>C</u> - <u>C</u>H₃
(b) sp³ sp²
H₃<u>C</u> - <u>C</u>H = <u>C</u>H₂
sp²
(c) sp³ sp
H₃<u>C</u> - <u>C</u> ≡ <u>C</u> - <u>C</u>H₂OH
sp sp³
(d) sp³ sp²
H₃<u>C</u> - <u>C</u>H=O
Explanation:
Alkanes or the carbons with all the single bonds are sp³ hybridized.
Alkenes or the carbons with double bond(s) are sp² hybridized.
Alkynes or the carbons with triple bond are sp hybridized.
Considering:
(a) H₃C-CH₃ , Both the carbons are bonded by single bond so both the carbons are sp³ hybridized.
Hence,
sp³ sp³
H₃<u>C</u> - <u>C</u>H₃
(b) H₃C-CH=CH₂ , The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp² hybridized because they are bonded by double bond.
Hence,
sp³ sp²
H₃<u>C</u> - <u>C</u>H = <u>C</u>H₂
sp²
(c) H₃C-C≡C-CH₂OH , The carbons of the methyl group and alcoholic group are sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp hybridized because they are bonded by triple bond.
Hence,
sp³ sp
H₃<u>C</u> - <u>C</u> ≡ <u>C</u> - <u>C</u>H₂OH
sp sp³
(d)CH₃CH=O, The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The other carbon is sp² hybridized because it is bonded by double bond to oxygen.
Hence,
sp³ sp²
H₃<u>C</u> - <u>C</u>H=O