CameraIcon

SearchIcon

MyQuestionIcon

1

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

Solubility Product

Ka1,Ka2 and ...

Question

$K_{a 1}, K_{a 2}$ and $K_{a 3}$ values for $H_{3} P O_{4}$ are $10^{- 3}, 10^{- 8}$ and $10^{- 12}$ respectively. If $K_{w} (H_{2} O) = 10^{- 14}$ , then which is correct?

A

Dissociation constant of

H P O_{4}^{2 -}

is

10^{- 12}

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

B

K_{b}

of

H P O_{4}^{2 -} 10^{- 6}

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

C

K_{b}

of

H_{2} P O_{4}^{-} 10^{- 11}

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

D

All are correct

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

Open in App

Solution

The correct option is C All are correct
$H_{3} P O_{4} ⇌ H^{+} + (H_{2} P O_{4})^{-}; K_{a 1} = 10^{- 3}$
$(H_{2} P O_{4})^{-} ⇌ H^{+} + (H P O_{4})^{2 -}$ ; $K_{a 2} = 10^{- 8}$
$(H P O_{4})^{- 2} ⇌ H^{+} + (P O_{4})^{- 3}$ ; $K_{a 3} = 10^{- 12}$
Conjugate acid $⇌ H^{+} +$ Conjugate base
For a reaction, $K_{a}$ is calculated for the conjugate acid and $K_{b}$ is calculated for the conjugate base.
$K_{w} = K_{a} . K_{b}$
So, for the first reaction, $K_{b}$ of $(H_{2} P O_{4})^{-} = \frac{10^{- 14}}{10^{- 3}} = 10^{- 11}$
For second reaction, $K_{b}$ of $(H P O_{4})^{2 -} = \frac{10^{- 14}}{10^{- 8}} = 10^{- 6}$
For reaction 3, $K_{b}$ of $(P O_{4})^{3 -} = \frac{10^{- 14}}{10^{- 12}} = 10^{- 2}$

Suggest Corrections

thumbs-up

0

Similar questions

K_{a_{1}}, K_{a_{2}}

K_{a_{3}}

H_{3} P O_{4}

10^{- 3}, 10^{- 6}

10^{- 12}

respectively. If

K_{w} (H_{2} O) = 10^{- 14}

, then:
What is

K_{b}

H P O_{4}^{-}

K_{a_{1}}, K_{a_{2}}, K_{a_{3}}

H_{3} {P O}_{4}

10^{- 3}, 10^{- 8}, 10^{- 12}

respectively. If

K_{w} (H_{2} O) = 10^{- 14}

, then
(i) What is dissociation constant of

H {P O}_{4}^{2 -}

K_{b}

H {P O}_{4}^{2 -}

K_{b}

H_{2} {P O}_{4}^{2 -}

?
(iv) What order of

K_{b}

{P O}_{4}^{3 -} (K_{b_{3}}), H {P O}_{4}^{2 -} (K_{b_{2}})

H_{2} {P O}_{4}^{-} (K_{b_{1}})

K_{a_{1}}, K_{a_{2}}

K_{a_{3}}

be the first, second and third ionization constant of

H_{3} P O_{4}

K_{a_{1}} >> K_{a_{2}} >> K_{a_{3}}

which is/ are correct

K_{a_{1}}, K_{a_{2}}

K_{a_{3}}

be the first, second and third ionization constant of

H_{3} P O_{4}

K_{a_{1}} ≫ K_{a_{2}} ≫ K_{a_{3}}

, then which of the following is/are correct?

Q. Calculate the pH of

0.1

M solution of (i)

N a H C O_{3}

N a_{2} H P O_{4}

N a H_{2} P O_{4}

C O_{2} + H_{2} O ⇌ H^{+} + C O_{3}^{2 -}

K_{1} = 4.2 \times 10^{- 7}

M

H C O_{3}^{-} ⇌ H^{+} + C O_{3}^{2 -}

K_{2} = 4.8 \times 10^{- 11}

M

H_{2} P O_{4} ⇌ H^{+} + H_{2} P O_{4}^{-}

K_{1} = 7.5 \times 10^{- 3}

M

H_{2} P O_{4}^{-} ⇌ H^{+} + H P O_{4}^{2 -}

K_{2} = 6.2 \times 10^{- 8}

M

H P O_{4}^{2 -} ⇌ H^{+} + P O_{4}^{3 -}

K_{3} = 1.0 \times 10^{- 12}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Solubility and Solubility Product

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Solubility Product

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon