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Ionic Equilibrium question

2022 · Shift 1 · Q3
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Ionic Equilibrium question

2022 · Shift 1 · Q3

JEE AdvancedChemistryIonic EquilibriumNumerical+3 / −1
A solution is prepared by mixing 0.01 mol0.01 \mathrm{~mol}0.01 mol each of H2CO3,NaHCO3,Na2CO3\mathrm{H}_{2} \mathrm{CO}_{3}, \mathrm{NaHCO}_{3}, \mathrm{Na}_{2} \mathrm{CO}_{3}H2​CO3​,NaHCO3​,Na2​CO3​, and NaOH\mathrm{NaOH}NaOH in 100 mL100 \mathrm{~mL}100 mL of water. pHp \mathrm{H}pH of the resulting solution is ‾\underline{\hspace{2cm}}​. [Given: p Ka1p \mathrm{~K}_{\mathrm{a} 1}p Ka1​ and p Ka2p \mathrm{~K}_{\mathrm{a} 2}p Ka2​ of H2CO3\mathrm{H}_{2} \mathrm{CO}_{3}H2​CO3​ are 6.376.376.37 and 10.32, respectively; log⁡2=0.30\log 2=0.30log2=0.30 ]
Numerical answer
View written solutionFree

Correct answer: 10.00TO10.04

  1. List the species mixed initially

In 100 mL100\,\text{mL}100mL water, we mix 0.010.010.01 mol each of:

  • H2CO3\mathrm{H_2CO_3}H2​CO3​
  • NaHCO3\mathrm{NaHCO_3}NaHCO3​
  • Na2CO3\mathrm{Na_2CO_3}Na2​CO3​
  • NaOH\mathrm{NaOH}NaOH

So initially:

  • H2CO3=0.01\mathrm{H_2CO_3} = 0.01H2​CO3​=0.01 mol
  • HCO3−=0.01\mathrm{HCO_3^-} = 0.01HCO3−​=0.01 mol
  • CO32−=0.01\mathrm{CO_3^{2-}} = 0.01CO32−​=0.01 mol
  • OH−=0.01\mathrm{OH^-} = 0.01OH−=0.01 mol
  1. First, react the strong base OH−\mathrm{OH^-}OH− completely

OH−\mathrm{OH^-}OH− will react with the acidic species H2CO3\mathrm{H_2CO_3}H2​CO3​:

H2CO3+OH−→HCO3−+H2O\mathrm{H_2CO_3 + OH^- \rightarrow HCO_3^- + H_2O}H2​CO3​+OH−→HCO3−​+H2​O

Since both are present in equal moles, 0.010.010.01 mol each, they completely consume each other.

After reaction:

  • H2CO3=0\mathrm{H_2CO_3} = 0H2​CO3​=0
  • OH−=0\mathrm{OH^-} = 0OH−=0
  • HCO3−\mathrm{HCO_3^-}HCO3−​ increases by 0.010.010.01 mol

So final moles become:

  • HCO3−=0.01+0.01=0.02\mathrm{HCO_3^-} = 0.01 + 0.01 = 0.02HCO3−​=0.01+0.01=0.02 mol
  • CO32−=0.01\mathrm{CO_3^{2-}} = 0.01CO32−​=0.01 mol

Thus the final solution contains only the buffer pair:

HCO3−/CO32−\mathrm{HCO_3^- / CO_3^{2-}}HCO3−​/CO32−​
  1. Use Henderson–Hasselbalch equation for the second dissociation

For the equilibrium

HCO3−⇌H++CO32−\mathrm{HCO_3^- \rightleftharpoons H^+ + CO_3^{2-}}HCO3−​⇌H++CO32−​

we use pKa2=10.32pK_{a2} = 10.32pKa2​=10.32.

The Henderson equation is:

pH = pK_{a2} + \log\left(\frac{[\mathrm{CO_3^{2-}}]}{[\mathrm{HCO_3^-}]}}\right)

Now,

[CO32−][HCO3−]=0.010.02=12\frac{[\mathrm{CO_3^{2-}}]}{[\mathrm{HCO_3^-}]} = \frac{0.01}{0.02} = \frac{1}{2}[HCO3−​][CO32−​]​=0.020.01​=21​

Hence,

pH=10.32+log⁡(12)pH = 10.32 + \log\left(\frac{1}{2}\right)pH=10.32+log(21​)

Using log⁡2=0.30\log 2 = 0.30log2=0.30,

log⁡(12)=−log⁡2=−0.30\log\left(\frac{1}{2}\right) = -\log 2 = -0.30log(21​)=−log2=−0.30

Therefore,

pH=10.32−0.30=10.02pH = 10.32 - 0.30 = 10.02pH=10.32−0.30=10.02
  1. Final answer
10.02\boxed{10.02}10.02​

This lies in the stored correct range 10.0010.0010.00 to 10.0410.0410.04.

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