Subject:
ChemistryAuthor:
cynthiarhodesCreated:
1 year ago⌬ According to the Question :
Given That-
[tex] \\ \longrightarrow\sf \: Ka = {a}^{2}c[/tex]
[tex] \\ [/tex]
★Need To Find-
[tex] \\ [/tex]
★Formula To Be Used-
[tex] \small\longrightarrow \sf \: \boxed{\bf\pink{a = \frac{Percent \: Dissociation}{100} }}[/tex]
[tex] \\ [/tex]
❒ Let's Find-
[tex] = \sf \: \frac{0.5}{100} \\ [/tex]
[tex] \sf➪ \: 5 × 10⁻⁴ \\ [/tex]
[tex] \sf➪ \: C = 0.02 M \\ [/tex]
[tex] \sf➪ \: 2 × 10⁻²M \\ [/tex]
[tex] \\ {\underline{\rule{160pt}{4pt}}} [/tex]
✠Hence, Kₐ
[tex] \\ [/tex]
[tex]\longmapsto \sf \: (5 \times 10⁻⁴) ^{2} \times 2 \times 10 ^{ - 2} \\ \\ \longmapsto \sf \: 25 \times 10⁻⁸ \times 2 \times {10}^{ - 2} \: \: \: \\ \\ \longmapsto \sf \: 50 \times 10 ^{ - 10} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longmapsto \sf \: \boxed{\bf\pink{ 5 \times {10}^{ - 9} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]
❛❛ Therefore, The Dissociation Constant Of Acid Is = 5 × 10⁻⁹ ❜❜
[tex] \\ {\underline{\rule{300pt}{9pt}}} [/tex]
Author:
sablestanton
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11[tex]{ \underline {\large{ \sf{Solution}}}}[/tex]
The dissociation constant of acid is Given by [tex] \rm{Ka=a^2c.}[/tex]
We Know,Putting Values
[tex] \dashrightarrow\rm{ = \dfrac{0.05}{100}}[/tex]
[tex] \dashrightarrow \rm{= 5 \times 10^{-4}}[/tex]
[tex]\dashrightarrow\rm{ \green{C = 0.02 M}}[/tex]
[tex] \dashrightarrow \rm{=2 \times 10^{-2} {m}}[/tex]
Hence, K =
[tex]{ \dashrightarrow{\rm{(5 \times 10^{ -4})²×2×10^{-²} }}}[/tex]
[tex] {\dashrightarrow \rm{25 × 10^{ - 8} × 2 × 10^{-2}}}[/tex]
[tex]{ \dashrightarrow\rm{=50\times 10^{-10}}}[/tex]
[tex]{ \green{ \dashrightarrow\rm{-5 × 10^{⁹}}}}[/tex]
Therefore,[tex]{ \underline{ \underline{ \rule{500pt}{5pt}}}}[/tex]
Author:
chelseamatthews
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