H2CO3 ionises as H2CO3- H- HCO3- K1- 4-3- 10-7 H2CO3- H- HCO32-K2 - 5-6- 10-11calculate pH value of 0-12 M solution of Na2CO3-

The correct option is A 11.66 2Na^+CO^2 _3 +H_2O nbsp;⇌ 2Na^+ + HCO^ _3 +OH^ K_h= C × (α)^2= K_w/K_a Substituting the given values we get α=0.0 ...

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

Byju's Answer

Standard XII

Chemistry

Salt of Strong Acid and Weak Base

H2CO3 ionises...

Question

$H_{2} C O_{3}$ ionises as
$H_{2} C O_{3} ⇌ H^{+} + H C O_{3}^{-} K_{1} = 4.3 \times 10^{- 7}$
$H_{2} C O_{3}^{-} ⇌ H^{+} + H C O_{3}^{2 -} K_{2} = 5.6 \times 10^{- 11}$
calculate pH value of 0.12 M solution of $N a_{2} C O_{3}$ .

11.66

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

6.11

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

1.166

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

0.1166

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

Open in App

Solution

The correct option is A 11.66
$2 N a^{+} C O_{3}^{2 -} + H_{2} O ⇌ 2 N a^{+} + H C O_{3}^{-} + O H^{-}$
$K_{h} = C \times (α)^{2} = \frac{K_{w}}{K_{a}}$
Substituting the given values we get $α = 0.0385$
$[O H^{-}] = C \times α = 4.269 \times 10^{- 3}$
pOH=2.3345 and pH=11.66.

Suggest Corrections

Similar questions

H_{2} {C O}_{3}

ionises as

H_{2} {C O}_{3} ⇌ H^{+} + H {C O}_{3}^{-}; K_{1} = 4.3 \times 10^{- 7}

H {C O}_{3}^{-} ⇌ H^{+} + {C O}_{3}^{2 -}; K_{2} = 5.6 \times 10^{- 11}

Calculate the degree of hydrolysis and pH value of

0.12 M

{N a}_{2} {C O}_{3}

solution.

Q. Find the concentration of

H^{+}

ions in an aqueous solution, which is saturated with

H_{2} S

(

0.1 M

) as well as

H_{2} C O_{3}

(

0.2 M

[Use data:

K_{1} = 10^{- 7}, K_{2} = 10^{- 14}

for

H_{2} S, K_{1} = 4 \times 10^{- 7}, K_{2} = 4 \times 10^{- 11}

for

H_{2} C O_{3}]

Q. Concentration of

H^{+}

ions in 0.1 M

H_{2} C O_{3}

(K_{1} = 4 \times 10^{- 7}, K_{2} = 4 \times 10^{- 11}) :

An aqueous solution of carbonic acid ( $H_{2} C O_{3}$ ) contains

Q. Calculate the pH of

0.1

M solution of (i)

N a H C O_{3}

, (ii)

N a_{2} H P O_{4}

and (iii)

N a H_{2} P O_{4}

Given that:

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

;

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

H C O_{3}^{-} ⇌ H^{+} + C O_{3}^{2 -}

;

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

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

;

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

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

;

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

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

;

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

Join BYJU'S Learning Program

H2CO3 ionises as H2CO3⇌H++HCO−3K1=4.3×10−7H2CO−3⇌H++HCO2−3K2=5.6×10−11calculate pH value of 0.12 M solution of Na2CO3.

The correct option is A 11.662Na+CO2−3+H2O⇌2Na++HCO−3+OH−Kh=C×(α)2=KwKaSubstituting the given values we get α=0.0385[OH−]=C×α=4.269×10−3pOH=2.3345 and pH=11.66.

$H_{2} C O_{3}$ ionises as
$H_{2} C O_{3} ⇌ H^{+} + H C O_{3}^{-} K_{1} = 4.3 \times 10^{- 7}$
$H_{2} C O_{3}^{-} ⇌ H^{+} + H C O_{3}^{2 -} K_{2} = 5.6 \times 10^{- 11}$
calculate pH value of 0.12 M solution of $N a_{2} C O_{3}$ .

The correct option is A 11.66
$2 N a^{+} C O_{3}^{2 -} + H_{2} O ⇌ 2 N a^{+} + H C O_{3}^{-} + O H^{-}$
$K_{h} = C \times (α)^{2} = \frac{K_{w}}{K_{a}}$
Substituting the given values we get $α = 0.0385$
$[O H^{-}] = C \times α = 4.269 \times 10^{- 3}$
pOH=2.3345 and pH=11.66.