If equilibrium constant for the reaction, XO -( aq )+ H 2 O (1) leftharpoons HXO ( aq )+ OH -( aq ) is 0.36 × 10-6 then find the value of dissociation constant ( K a ) for HX O ?

Question

If equilibrium constant for the reaction, XO -( aq )+ H 2 O (1) leftharpoons HXO ( aq )+ OH -( aq ) is 0.36 × 10-6 then find the value of dissociation constant ( K a ) for HX O ? (A) 0.36 × 10-8 (B) 2.8 × 10-8 (C) 2.8 × 10-10 (D) 0.36 × 10-6

Tardigrade Admin · Accepted Answer

HXO leftharpoons H ++ XO - K a =([ H +][ XO ]/[ HXO ]) XO -+ H 2 O leftharpoons HXO + OH - K eq =([ HXO ][ OH -]/[ XO -]) K a ⋅ K eq =[ H +][ OH -] = K w =1 × 10-4 K a =( K w / K eq )=(1.0 × 10-14/0.36 × 10-6) =2.8 × 10-8

Q. If equilibrium constant for the reaction,
$X O^{-} (a q) + H_{2} O (1) ⇌ H XO (a q) + O H^{-} (a q)$
is $0.36 \times 1 0^{- 6}$ then find the value of dissociation constant $(K_{a})$ for $H XO$ ?

$0.36 \times 1 0^{- 8}$

$2.8 \times 1 0^{- 8}$

$2.8 \times 1 0^{- 10}$

$0.36 \times 1 0^{- 6}$

Q. If equilibrium constant for the reaction, XO−(aq)+H2​O(1)⇌HXO(aq)+OH−(aq) is 0.36×10−6 then find the value of dissociation constant (Ka​) for HXO ?

0.36×10−8

2.8×10−8

2.8×10−10

0.36×10−6

HXO⇌H++XO− Ka​=[HXO][H+][XO]​ XO−+H2​O⇌HXO+OH− Keq​=[XO−][HXO][OH−]​ Ka​⋅Keq​=[H+][OH−] =Kw​=1×10−4 Ka​=Keq​Kw​​=0.36×10−61.0×10−14​ =2.8×10−8

Q. If equilibrium constant for the reaction,
$X O^{-} (a q) + H_{2} O (1) ⇌ H XO (a q) + O H^{-} (a q)$
is $0.36 \times 1 0^{- 6}$ then find the value of dissociation constant $(K_{a})$ for $H XO$ ?

$0.36 \times 1 0^{- 8}$

$2.8 \times 1 0^{- 8}$

$2.8 \times 1 0^{- 10}$

$0.36 \times 1 0^{- 6}$