Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Calculate approximate pH of the resultant solution formed by titration of 25mL of 0.04 M Na₂CO₃ with 50mL of 0.025M HCl [Given : pK_a1 = 6.4 and pK_a2 = 10.3; for H₂CO₃]

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

6.4

5.92

5.88

6.88

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

	Na₂CO₃	+ HCl →	NaHCO₃	+ NaCl
milli moles	= 1	= 1.25
	1-1=0	1.25-1 = 0.25	1	1

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 moles	1	0.25	0	0
	1 - 0.25 = 0.75	0.25 - 0.25 = 0	0.25	0

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

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

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!