Sulphurous acid H2SO3 has Ka1 - 1-7 - 10-2 and Ka2 - 6-4 - 10-8- The pH of 0-588 M H2SO3 is-Round off to the nearest integer

H_2 SO_3 (aq) amp; ⇌ amp;HSO^ _3 (aq) amp; + amp; H^+ (aq) 0.588 M = C amp; amp;C α_1 amp; amp;(C α_1 + C α_1 α_2) HSO^ _3 (aq) amp; ...

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Byju's Answer

Standard XII

Chemistry

Conjugate Acids and Bases

Sulphurous ac...

Question

Sulphurous acid $(H_{2} S O_{3})$ has $K_{a_{1}} = 1.7 \times 10^{- 2}$ and $K_{a_{2}} = 6.4 \times 10^{- 8}$ . The $p H$ of $0.588 M H_{2} S O_{3}$ is.
(Round off to the nearest integer)

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Solution

$\begin{matrix} H_{2} S O_{3} (a q) & ⇌ & H S O_{3}^{-} (a q) & + & H^{+} (a q) 0.588 M = C & C α_{1} & (C α_{1} + C α_{1} α_{2}) \end{matrix}$

$\begin{matrix} H S O_{3}^{-} (a q) & ⇌ & S O_{3}^{2 -} (a q) & + & H^{+} (a q) C α_{1} (1 - α_{2}) & C α_{1} α_{2} & (C α_{1} + C α_{1} α_{2}) \end{matrix}$

$α_{1} = \sqrt{\frac{1.7 \times 10^{- 2}}{0.588}} = \sqrt{\frac{17}{289 \times 2}}$

Therefore, $α_{1} << 1 \Rightarrow (1 - α_{1}) ≃ 1$
Hence, $α_{2} << 1 \Rightarrow (1 - α_{2}) \approx 1$
$∴ [H^{+}] = C α_{1}$
$= \sqrt{K_{a_{1}} \times C} = \sqrt{17 \times 10^{- 3} \times 0.588}$
$= 0.099$

$p H = - {log}_{10} 0.099 = 1$
Nearest integer $= 1$

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Similar questions

Q. Sulphurous acid

(H_{2} S O_{3})

has

K_{a_{1}} = 1.7 \times 10^{- 2}

and

K_{a_{2}} = 6.4 \times 10^{- 8}

. The

p H

0.588 M H_{2} S O_{3}

is.
(Round off to the nearest integer)

Q. What is the

p H

0.01 M

glycine solution?
For glycine

K_{a_{1}} = 4.5 \times 10^{- 3}

and

K_{a_{2}} = 1.7 \times 10^{- 10}

298 K

Q. Selenious acid

(H_{2} S e O_{3})

, a diprotic acid has

K_{a_{1}} = 3.0 \times 10^{- 3}

and

K_{a_{2}} = 5.0 \times 10^{- 8}

. What is the

[O H^{-}]

of a

0.30 M

solution of a seleniuos acid?

Q. Assertion :When

0.1 M

weak diprotic acid

H_{2} A

dissociates with its dissociation constants

K_{a_{1}} = 10^{- 3}

and

K_{a_{2}} = 10^{- 8}

, then

[A^{2 -}]

is almost equal to

10^{- 3} M

Reason: Since

K_{a_{2}} ≪ K_{a_{1}}

for

0.1 M H_{2} A

, so

[A^{2 -}]

is negligible w.r.t.

[H A^{-}]

Q. Approximate

p H

0.01 M N a H A

is calculated by ?

(K_{a_{1}} = 10^{- 6}

and

K_{a_{2}} = 10^{- 8}

are ionization constants of

H_{2} A)

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Sulphurous acid (H2SO3) has Ka1=1.7×10−2 and Ka2=6.4×10−8. The pH of 0.588 M H2SO3 is. (Round off to the nearest integer)

Sulphurous acid $(H_{2} S O_{3})$ has $K_{a_{1}} = 1.7 \times 10^{- 2}$ and $K_{a_{2}} = 6.4 \times 10^{- 8}$ . The $p H$ of $0.588 M H_{2} S O_{3}$ is.
(Round off to the nearest integer)