You visited us 3 times! Enjoying our articles? Unlock Full Access!

Solubility Product

The pH of a 0...

Question

The pH of a 0.001 M solution of $N H_{4} O H$ (having dissociation constant $K_{b}$ $= 1.0 \times 10^{- 5}$ ) is equal to:

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

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

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

Open in App

Solution

The correct option is D 10
$N H_{4} O H \to N H_{4}^{+} + O H^{-}$

Conc. of $[O H^{-}] = c α = \sqrt{C \times K_{b}} = \sqrt{0.001 \times 10^{- 5}} = 10^{- 4}$

So $p O H = - l o g [O H^{-}] = 4$

$p H = 14 - p O H = 10$

Suggest Corrections

Similar questions

p H

0.1 M

{N H}_{4} O H

K_{b} = 1.0 \times 10^{- 5})

N H_{4} O H

p H

K_{b}

N H_{4} O H

1.8 \times 10^{- 5} M

{N H}_{4} O H

{N H}_{4} C I .

[{O H}^{-}]

[K_{b} = 2 \times 10^{- 5}]

C H_{3} C O O H

N H_{4} O H

10^{- 5}

C H_{3} C O O H

3

N H_{4} O H

p H

0.1 M

{C H}_{3} C O O {N H}_{4}

{C H}_{3} C O O H

{N H}_{4} O H

K_{a} = 1.8 \times 10^{- 5}

K_{b} = 1.8 \times 10^{- 5}

Join BYJU'S Learning Program