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

Salt of Weak Acid and Weak Base

First ionizat...

Question

First ionization of phosphoric acid is:
$H_{3} {P O}_{4} ⇌ H_{3} {P O}_{4}^{-} + H^{+}$
$p K_{a_{1}} = 2.21$
The dissociation constant of conjugate base of $H_{3} {P O}_{4}$ will be:

6.17 \times 10^{- 8}

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

1.62 \times 10^{- 12}

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

3.48 \times 10^{- 11}

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

4.62 \times 10^{- 2}

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

Open in App

Solution

The correct option is B $1.62 \times 10^{- 12}$
Conjugate base of $H_{3} {P O}_{4}$ is $H_{2} {P O}_{4}^{-}$ .
$H_{2} {P O}_{4}^{-} ⇌ {H P O}_{4}^{2 -} + H^{+}$
So, it dissociation constant will be equal to ${K a}_{2}$ of $H_{3} {P O}_{4}$ .
now, ${p K a}_{2} = 7.21 \Rightarrow - l o g {K a}_{2} = 7.21$
$∴$ ${K a}_{2} = 10^{- 7.21} = 6.16 \times 10^{- 8}$

Suggest Corrections

Similar questions

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

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^{- 7}

H P O_{4}^{- 2} ⇌ H^{+} + {P O}_{4}^{- 3}; K_{a_{3}} = 10^{- 13}

0.1 M

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

{P O}_{4}^{3 -}

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

[H^{+}]

H_{3} P O_{4}

H_{3} {P O}_{4} ⇌ H^{+} + H_{2} {P O}_{4}^{-} {K a}_{1} = 10^{- 4}

H_{2} {P O}_{4}^{-} ⇌ H^{+} + H {P O}_{4}^{- 2} {K a}_{2} = 10^{- 9}

H {P O}_{4}^{- 2} ⇌ H^{+} + {P O}_{4}^{3 -} {K a}_{3} = 10^{- 13}

p H

0.1 M

K_{3} {P O}_{4}

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

1.3 \times 10^{- 12}

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

H_{3} {P O}_{4} + H_{2} O \to H_{3} O^{+} + H_{2} {P O}_{4}^{-}

H_{2} {P O}_{4}^{-} + H_{2} O \to H {P O}_{4}^{2 -} + H_{3} O^{+}

H_{2} {P O}_{4}^{-} + {O H}^{-} \to H_{3} {P O}_{4} + O^{2 -}

Join BYJU'S Learning Program