PH of 0-1 M NaHCO3 if K1-4-5-10-7-K2-4-5-10-11-

The correct option is C 9.6NaHCO3 + H2O lt;— gt; H2CO3 + NaOH #160;Kb= kw/ka #160; #160;Where ka is equal to K1=4.5 x 10^ 7 #160; #160;So kb=(1 ...

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

Byju's Answer

Standard VII

Chemistry

pH Scale

pH of 0.1 M ...

Question

pH of 0.1 M $N a H C O_{3}$ if $K_{1} = 4.5 \times 10^{- 7}, K_{2} = 4.5 \times 10^{- 11}$ .

8.34

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

9.6

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

8.2

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

9.9

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

Open in App

Solution

The correct option is C 9.6
NaHCO3 + H2O <---> H2CO3 + NaOH

Kb= kw/ka

Where ka is equal to K1=4.5 x 10^ -7

So kb=(10^-14)/(4.5*10^-7) is equal to 2.22 x 10^-8

Dissociation is low 2.22 x 10^8 = [OH] ^ 2/0.1

[OH] is equal to 4.7 x 10 ^ -5

POH = -log[OH] = 4.32

So pH = 14-4.32=9.68

Suggest Corrections

Similar questions

Q. What is the

p H

0.1 M

N a H {C O}_{3}

(Given

K_{1} = 4.5 \times 10^{- 7}, K_{2} = 4.5 \times 10^{- 11}

for carbonic acids.)

Q. Calculate the

p H

(nearest integer) of

0.010 M

N a H {C O}_{3}

solution.

K_{1} = 4.5 \times 10^{- 7}

and

K_{2} = 4.7 \times 10^{- 11}

for carbonic acid.

Q. What is the

p H

0.01 M N a H C O_{3}

?
Given :

K_{a_{1}} a n d K_{a_{2}}

for

H_{2} C O_{3}

are

4.5 \times 10^{- 7} a n d 4.7 \times 10^{- 11}

respectively at

25^{o} C

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}

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}) :

Join BYJU'S Learning Program

pH of 0.1 M NaHCO3 if K1=4.5×10−7,K2=4.5×10−11.

The correct option is C 9.6NaHCO3 + H2O <---> H2CO3 + NaOH Kb= kw/ka Where ka is equal to K1=4.5 x 10^ -7 So kb=(10^-14)/(4.5*10^-7) is equal to 2.22 x 10^-8 Dissociation is low 2.22 x 10^8 = [OH] ^ 2/0.1 [OH] is equal to 4.7 x 10 ^ -5 POH = -log[OH] = 4.32 So pH = 14-4.32=9.68

pH of 0.1 M $N a H C O_{3}$ if $K_{1} = 4.5 \times 10^{- 7}, K_{2} = 4.5 \times 10^{- 11}$ .

The correct option is C 9.6
NaHCO3 + H2O <---> H2CO3 + NaOH

Kb= kw/ka

Where ka is equal to K1=4.5 x 10^ -7

So kb=(10^-14)/(4.5*10^-7) is equal to 2.22 x 10^-8

Dissociation is low 2.22 x 10^8 = [OH] ^ 2/0.1

[OH] is equal to 4.7 x 10 ^ -5

POH = -log[OH] = 4.32

So pH = 14-4.32=9.68