linandrew41
Answered

Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

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 :

DᴀʀᴋPᴀʀᴀᴅᴏx

[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]

Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.