$p H$ of $0.08 m o l d m^{- 3} H O C l$ solution is $2.85$ . Calculate its ionization constant.

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Solution

$H O C l$ is a weak acid.

$p H = - l o g [H^{+}]$

$[H^{+}] = \sqrt{K_{a} \times C}$

To find $K_{a}$ for weak mono basic acid

$p H = 2.85$

$[H^{+}] = 10^{- p H}$

$= 10^{- 2.85}$

$= 1.41 \times 10^{- 3}$

$∴ K_{a} = \frac{[H^{+}]^{2}}{C}$

$= \frac{(1.41 \times 10^{- 3})^{2}}{0.08}$

$= \frac{1.99 \times 10^{- 6}}{0.08}$

$= 0.249 \times 10^{- 4}$

$= 2.49 \times 10^{- 5}$

Therefore, the ionization constant is $2.49 \times 10^{- 5} m o l e / l i t e r$ .

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Q. pH of 0.08 mol dm-3 HOCI solution is 2.85.lonisation constant of HOCI is (antilog of 0.151.413)

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pH of 0.08 mol dm−3 HOCl solution is 2.85. Calculate its ionization constant.

HOCl is a weak acid. pH=−log[H+] [H+]=√Ka× C To find Ka for weak mono basic acid pH=2.85 [H+]=10−pH =10−2.85 =1.41×10−3 ∴ Ka=[H+]2C =(1.41×10−3)20.08 =1.99×10−60.08 =0.249×10−4 =2.49×10−5 Therefore, the ionization constant is 2.49×10−5 mole/liter.

$p H$ of $0.08 m o l d m^{- 3} H O C l$ solution is $2.85$ . Calculate its ionization constant.