What is the pH of 0-01 M NaHCO3 -Given - Ka1 and Ka2 for H2CO3 are 4-5 - 10-7 and 4-7- 10-11 respectively at 25 oC-

PK_a1= log(K_a1) pK_a1= log(4.5× 10^ 7)=6.34 pK_a2= log(K_a2) pK_a2= log(4.7× 10^ 11)=10.32 For an amphiprotic ion pH is calculated as: pH= 12[pK_a_1+pK_a_2] ...

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Standard XI

Chemistry

Introduction and Degree of Hydrolysis

What is the p...

Question

What is the $p H$ of $0.01 M N a H C O_{3}$ ?
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 $25^{o} C$ .

5.66

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8.33

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10.57

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6.33

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Solution

The correct option is B 8.33
$p K_{a 1} = - log (K_{a 1}) p K_{a 1} = - log (4.5 \times 10^{- 7}) = 6.34$

$p K_{a 2} = - log (K_{a 2}) p K_{a 2} = - log (4.7 \times 10^{- 11}) = 10.32$

For an amphiprotic ion pH is calculated as:
$p H = \frac{1}{2} [p K_{a_{1}} + p K_{a_{2}}] p H = \frac{1}{2} [6.34 + 10.32] = 8.33$

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Introduction and Degree of Hydrolysis

Standard XI Chemistry

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What is the pH of 0.01 M NaHCO3 ? Given : Ka1 and Ka2 for H2CO3 are 4.5×10−7 and 4.7×10−11 respectively at 25 oC.

The correct option is B 8.33pKa1=−log(Ka1)pKa1=−log(4.5×10−7)=6.34 pKa2=−log(Ka2)pKa2=−log(4.7×10−11)=10.32 For an amphiprotic ion pH is calculated as: pH=12[pKa1+pKa2]pH=12[6.34+10.32]=8.33

What is the $p H$ of $0.01 M N a H C O_{3}$ ?
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 $25^{o} C$ .