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Q.

Calculate approximate pH of the resultant solution formed by titration of 25mL of 0.04 M Na2CO3 with 50mL of 0.025M HCl [Given : pKa1 = 6.4 and pKa2 = 10.3; for H2CO3]

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a

6.4

b

5.92

c

5.88

d

6.88

answer is B.

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Detailed Solution

 

 Na2CO3+        HCl     →NaHCO3+     NaCl
milli moles= 1= 1.25  
 1-1=01.25-1 = 0.2511

Na2CO3 is the limiting reagent and so all of will be converted to NaHCO3 which will further react with the leftover HCl in the following reaction.

 

 NaHCO3+        HCl     →H2CO3+     NaCl
milli moles10.2500
 1 - 0.25 = 0.750.25 - 0.25 = 00.250

The resultant solution contains a weak diprotic acid H2CO3 and an incomplete amphiprotic salt NaHCO3. The amphiprotic substance 

\left( {HCO_3^ - } \right)

will not show its acidic behaviour as there is a stronger acid H2CO3 is present.

 

The system is a buffer

{P^H} = {\text{ }}{P^{{K_{{a_1}}}}} + \log \frac{{\left[ {HCO_3^ - } \right]}}{{\left[ {{H_2}C{O_3}} \right]}}
\Rightarrow {P^H} = 6.4 + \log \frac{{0.75}}{{0.25}} = 6.87

 

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