An acid HA ionises as HA leftharpoons H+ + A- The pH of 1.0 M solution is 5. Its dissociation constant would be:

Question

An acid HA ionises as HA leftharpoons H+ + A- The pH of 1.0 M solution is 5. Its dissociation constant would be: (A) 5 (B) 5 × 10-8 (C) 1 × 10-5 (D) 1 × 10-10

Tardigrade Admin · Accepted Answer

Ka = ([H+)2/C-[H+)), [H+] < < C = ((10-5)2/1) = 10-10

Q. An acid HA ionises as
$H A ⇌ H^{+} + A^{-}$
The pH of 1.0 M solution is 5. Its dissociation constant would be:

5

$5 \times 1 0^{- 8}$

$1 \times 1 0^{- 5}$

$1 \times 1 0^{- 10}$

$K_{a} = \frac{[ H ^{+} ) ^{2}}{C - [ H ^{+} )}, [H^{+}] << C$
$= \frac{( 1 0 ^{- 5} ) ^{2}}{1} = 1 0^{- 10}$

Q. An acid HA ionises as HA⇌H++A− The pH of 1.0 M solution is 5. Its dissociation constant would be:

5

5×10−8

1×10−5

1×10−10

Ka​=C−[H+)[H+)2​,[H+]<<C =1(10−5)2​=10−10

Q. An acid HA ionises as
$H A ⇌ H^{+} + A^{-}$
The pH of 1.0 M solution is 5. Its dissociation constant would be:

$5 \times 1 0^{- 8}$

$1 \times 1 0^{- 5}$

$1 \times 1 0^{- 10}$

$K_{a} = \frac{[ H ^{+} ) ^{2}}{C - [ H ^{+} )}, [H^{+}] << C$
$= \frac{( 1 0 ^{- 5} ) ^{2}}{1} = 1 0^{- 10}$