3 years ago

Determing Valence Electrons

Chemistry

1 answer:

ioda3 years ago

6 0

Gallium ( Ga ) is the answer

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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.

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${ \qquad\qquad\huge\underline{{\sf Answer}}}$

Let's solve ~

Initial concentration of weak acid HA = 0.200 M

and dissociation constant ( ${ \alpha}$ ) is :

$\qquad \sf \dashrightarrow \: \alpha = \frac{dissociation \: \: percentage}{100}$

$\qquad \sf \dashrightarrow \: \alpha = \frac{9.4}{100} = 0.094$

Now, at initial stage :

$\textsf{ Conc of HA = 0.200 M}$

$\textsf{Conc of H+ = 0 M}$

$\textsf{Conc of A - = 0 M}$

At equilibrium :

$\textsf{Conc of HA = 0.200 - 0.094(0.200) = 0.200(1 - 0.094) = 0.200(0.906) = 0.1812 M}$

$\textsf{Conc of H+ = 0.094(0.200) = 0.0188 M}$

$\textsf{Conc of A - = 0.094(0.200) = 0.0188 M}$

Now, we know :

$\qquad \sf \dashrightarrow \: { K_a = \dfrac{[H+] [A-]}{[HA]}}$

( big brackets represents concentration )

$\qquad \sf \dashrightarrow \: { K_a = \dfrac{0.0188×0.0188}{0.1812}}$

$\qquad \sf \dashrightarrow \: { K_a = \dfrac{0.00035344}{0.1812}}$

$\qquad \sf \dashrightarrow \: { K_a \approx 0.00195 }$

$\qquad \sf \dashrightarrow \: {K_a \approx 1.9 × {10}^{-3} }$

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