K3 for ascorbic acid HAsc is 5 - 10-5-Calculate the -H- and - hydrolysis is an aq- Sol in which conc- of Asc- is 0-02 M-

The aq, solution contains a salt of strong base and weak acid. (Ascorbic Acid) K_n× K_a=K_w K_n=10^ 145× 10^ 15=2× 10^ 10 Hydrolysis Constant ⟶Ch^ ...

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Byju's Answer

Standard XII

Chemistry

Ostwald Dilution Law Mathematical Interpretation

K3 for ascorb...

Question

$K_{3}$ for ascorbic acid (HAsc) is $5 \times 10^{- 5}$ .Calculate the $[H^{+}]$ and $%$ hydrolysis is an aq. Sol in which conc. of $A s c^{-}$ is 0.02 M.

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Solution

The aq, solution contains a salt of strong base and weak acid. (Ascorbic Acid)
$K_{n} \times K_{a} = K_{w}$
$K_{n} = \frac{10^{- 14}}{5 \times 10^{- 15}} = 2 \times 10^{- 10}$ Hydrolysis Constant
$⟶ \frac{C h^{2}}{1 - h} = 2 \times 10^{- 10}$
neglecting $h$ , we get,
$C h^{2} = 2 \times 10^{- 10}$
$h = \sqrt{\frac{2 \times 10^{- 10}}{0.02}}$
$h = 10^{- 4}$
$%$ of hydrolysis $= 0.01 %$
Amount of $H^{+}$ ions $= C h = 0.02 \times 10^{- 4} = 2 \times 10^{- 2} m o l L^{- 1}$

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K3 for ascorbic acid (HAsc) is 5×10−5.Calculate the [H+] and % hydrolysis is an aq. Sol in which conc. of Asc− is 0.02 M.

The aq, solution contains a salt of strong base and weak acid. (Ascorbic Acid)Kn×Ka=KwKn=10−145×10−15=2×10−10 Hydrolysis Constant⟶Ch21−h=2×10−10neglecting h, we get,Ch2=2×10−10h=√2×10−100.02h=10−4% of hydrolysis=0.01%Amount of H+ ions=Ch=0.02×10−4=2×10−2molL−1

$K_{3}$ for ascorbic acid (HAsc) is $5 \times 10^{- 5}$ .Calculate the $[H^{+}]$ and $%$ hydrolysis is an aq. Sol in which conc. of $A s c^{-}$ is 0.02 M.