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Sagot :
[tex]{ \qquad\qquad\huge\underline{{\sf Answer}}} [/tex]
Let's solve ~
Initial concentration of weak acid HA = 0.200 M
and dissociation constant ([tex]{ \alpha} [/tex]) is :
[tex]\qquad \sf \dashrightarrow \: \alpha = \frac{dissociation \: \: percentage}{100} [/tex]
[tex]\qquad \sf \dashrightarrow \: \alpha = \frac{9.4}{100} = 0.094[/tex]
Now, at initial stage :
- [tex] \textsf{ Conc of HA = 0.200 M} [/tex]
- [tex] \textsf{Conc of H+ = 0 M} [/tex]
- [tex] \textsf{Conc of A - = 0 M} [/tex]
At equilibrium :
- [tex] \textsf{Conc of HA = 0.200 - 0.094(0.200) = 0.200(1 - 0.094) = 0.200(0.906) = 0.1812 M} [/tex]
- [tex] \textsf{Conc of H+ = 0.094(0.200) = 0.0188 M} [/tex]
- [tex] \textsf{Conc of A - = 0.094(0.200) = 0.0188 M} [/tex]
Now, we know :
[tex]\qquad \sf \dashrightarrow \: { K_a = \dfrac{[H+] [A-]}{[HA]}} [/tex]
( big brackets represents concentration )
[tex]\qquad \sf \dashrightarrow \: { K_a = \dfrac{0.0188×0.0188}{0.1812}} [/tex]
[tex]\qquad \sf \dashrightarrow \: { K_a = \dfrac{0.00035344}{0.1812}} [/tex]
[tex]\qquad \sf \dashrightarrow \: { K_a \approx 0.00195 } [/tex]
[tex]\qquad \sf \dashrightarrow \: {K_a \approx 1.9 × {10}^{-3} } [/tex]
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