A solution is prepared by mixing 0.01 mol each of H 2 CO 3 , NaHCO 3 , Na 2 CO 3 , and NaOH in 100 mL of water. p H of the resulting solution is . [0pt] [Given: p Ka 1 and p Ka 2 of H 2 CO 3 are overline 6.37 and 10.32 , respectively; log 2=0.30 ]

Chemistry · Ionic Equilibrium

JEE Advanced 2022 — Paper 1 — Question 37

A solution is prepared by mixing 0.01 mol each of

$\mathrm{H}_{2} \mathrm{CO}_{3}, \mathrm{NaHCO}_{3}, \mathrm{Na}_{2} \mathrm{CO}_{3}$ ,

and NaOH in 100 mL of water. $p \mathrm{H}$ of the resulting solution is $\qquad$ .

[0pt] [Given: $p \mathrm{Ka}_{1}$ and $p \mathrm{Ka}_{2}$ of $\mathrm{H}_{2} \mathrm{CO}_{3}$

are $\overline{6.37}$ and 10.32 , respectively; $\log 2=0.30$ ]

Answer: 10.02

Numerical answer — enter this value.

Step-by-step solution

& $\mathrm{H}_{2} \mathrm{CO}_{3}+\mathrm{NaOH} \longrightarrow$ &

$\mathrm{NaHCO}_{3}+\mathrm{H}_{2} \mathrm{O}$

& Initial mol & 0.01 & 0.01 & 0.01 After neutralization & 0 & 0 & $0.01+0.01$ & & & $=0.02$

Now, mixture contains 0.01 mole $\mathrm{Na}_{2} \mathrm{CO}_{3}$ and 0.02 mol of

$\mathrm{NaHCO}_{3}$ so it is a buffer.

$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}_{2}}+\log \frac{\left[\mathrm{CO}_{3}^{2-}\right]}{\left[\mathrm{HCO}_{3}^{-}\right]}$ $\mathrm{pH}=10.32+\log \frac{0.01}{0.02}$

$=10.32+\log \frac{1}{2}$ $=10.32-0.30$ $=10.02$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2022
Paper: Paper 1
Subject: Chemistry
Chapter: Ionic Equilibrium
Topic: Solutions with mixture of acids or bases