Answer
$Ka = 3.469\times 10^{- 8}$
$pKa = 7.46$
- Yes, the values on the table agree with the calculations.
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [OCl^-] = x$
-$[HOCl] = [HOCl]_{initial} - x$
2. Calculate the hydronium concentration
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 4.23}$
$[H_3O^+] = 5.888 \times 10^{- 5}$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][OCl^-]}{ [HOCl]}$
$Ka = \frac{x^2}{[InitialHOCl] - x}$
$Ka = \frac{( 5.888\times 10^{- 5})^2}{ 0.1- 5.888\times 10^{- 5}}$
$Ka = \frac{ 3.467\times 10^{- 9}}{ 0.09994}$
$Ka = 3.469\times 10^{- 8}$
4. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 3.469 \times 10^{- 8})$
$pKa = 7.46$
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