If the ionic product of water at 40oC is 4 - 10-14- The concentration of OH- ions in water at 40-C is-

H_2O (l)⇌ H^+ (aq.)+OH^ (aq.) K_w=[H^+][OH^ ] 4×10^ 14=[H^+]×[OH^ ] nbsp; At equilibrium, [H^+]= [OH^ ] ∴ K_w=[OH^ ]^2 [OH^ ]=√(K_w)=√(4×10^ 14) [OH^ ]=2×1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

Chemistry

Factors Affecting Alpha

If the ionic ...

Question

If the ionic product of water at $40^{o} C$ is $4 \times 10^{- 14}$ . The concentration of $O H^{-}$ ions in water at $40^{\circ} C$ is:

2 \times 10^{- 9} M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 \times 10^{- 7} M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 \times 10^{- 3} M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 \times 10^{- 7} M

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $2 \times 10^{- 7} M$
$H_{2} O (l) ⇌ H^{+} (a q .) + O H^{-} (a q .) K_{w} = [H^{+}] [O H^{-}]$
$4 \times 10^{- 14} = [H^{+}] \times [O H^{-}]$
At equilibrium,
$[H^{+}] = [O H^{-}] ∴ K_{w} = [O H^{-}]^{2} [O H^{-}] = \sqrt{K_{w}} = \sqrt{4 \times 10^{- 14}} [O H^{-}] = 2 \times 10^{- 7} M$

Suggest Corrections

Similar questions

Q. If the ionic product of water at

40^{o} C

4 \times 10^{- 14}

. The concentration of

O H^{-}

ions in water at

40^{\circ} C

is:

Q. The ionic product of water is

10^{- 14}

, What is the

H^{+}

ion concentration of a 0.1 M Na OH solution?

Q. Specific conductance of pure water at

25^{\circ}

0.58 \times 10^{- 7} m h o c m^{- 1}

. The ionic product of water (

K_{W}

) if ionic conductances of

H^{+}

and

O H^{-}

ions at infinite dilution are 350 and

198 m h o c m^{2}

respectively at

25^{\circ} C

is:

Q. The concentration of

[H^{+}]

and concentration of

[O H^{+}]

of a

0.1 M

aqueous solution of

2 %

ionised weak acid is [ionic product of water

= 1 \times 10^{- 14}

]

Q. The concentration of

[H^{+}]

and concentration of

[O H^{-}]

of a 0.1M aqueous solution of 2% ionised weak acid is:
[Ionic product of water

=

1 \times 10^{- 14}

]

Join BYJU'S Learning Program

Explore more

Factors Affecting Alpha

Standard XIII Chemistry

Join BYJU'S Learning Program

If the ionic product of water at 40oC is 4×10−14. The concentration of OH− ions in water at 40∘C is:

The correct option is D 2×10−7 MH2O (l)⇌H+ (aq.)+OH− (aq.)Kw=[H+][OH−] 4×10−14=[H+]×[OH−] At equilibrium, [H+]=[OH−]∴Kw=[OH−]2[OH−]=√Kw=√4×10−14[OH−]=2×10−7 M

If the ionic product of water at $40^{o} C$ is $4 \times 10^{- 14}$ . The concentration of $O H^{-}$ ions in water at $40^{\circ} C$ is: