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QJEEAdv22Ch3b90: JEE Advanced 2022 Chemistry - Class 11 JEEAdv22Ch3b90 Solution: A solution is prepared by mixing each of , and in of water. of…

A solution is prepared by mixing 0.01 mol0.01 \mathrm{~mol} each of H2CO3,NaHCO3,Na2CO3\mathrm{H}_{2} \mathrm{CO}_{3}, \mathrm{NaHCO}_{3}, \mathrm{Na}_{2} \mathrm{CO}_{3} , and NaOH\mathrm{NaOH} in 100 mL100 \mathrm{~mL} of water. pHp \mathrm{H} of the resulting solution is _________.

[Given: p Ka1p \mathrm{~K}_{\mathrm{a} 1} and p Ka2p \mathrm{~K}_{\mathrm{a} 2} of H2CO3\mathrm{H}_{2} \mathrm{CO}_{3} are 6.376.37 and 10.32, respectively; log⁡2=0.30\log 2=0.30 ]

Step-by-Step Verified Solution

First acid base reaction between H2CO3 and NaOH takes place.

JEE Advanced 2022 Paper 1 Online Chemistry - Ionic Equilibrium Question 6 English Explanation

In the final solution, we have 0.01 mole Na2CO3 and 0.02 moles of NaHCO3.

Here, we have a buffer of NaHCO3 and Na2CO3.

∴ pH=pKa2+log⁡[Salt][Acid]

=10.32+log⁡(0.010.1)(0.020.1)

=10.32+log⁡12

=10.32−log⁡2

=10.32−0.3

=10.02

∴ pH=10.02

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