Linandrew41idn
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A 0.200 M solution of a weak acid, HA, is 9.4% ionized. Using this information, calculate Ka for HA.

Select one:

a.
9.4 × 10−3


b.
1.8 × 10−3


c.
1.9 × 10−2


d.
3.8 × 10−3

Sagot :

DᴀʀᴋPᴀʀᴀᴅᴏxidn

[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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