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 ...

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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

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9.6

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8.2

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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

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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}) :

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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