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Ionic Equilibrium question

2023 · 13 Apr · Shift 2 · Q20
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  5. /2023 · 13 Apr · Shift 2 · Q20

Ionic Equilibrium question

2023 · 13 Apr · Shift 2 · Q20

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
20 mL of 0.1 M NaOH0.1 ~\mathrm{M} ~\mathrm{NaOH}0.1 M NaOH is added to 50 mL50 \mathrm{~mL}50 mL of 0.1 M0.1 ~\mathrm{M}0.1 M acetic acid solution. The pH\mathrm{pH}pH of the resulting solution is ‾\underline{\hspace{2cm}}​×10−2\times 10^{-2}×10−2(Nearest integer) Given : pKa(CH3COOH)=4.76\mathrm{pKa}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=4.76pKa(CH3​COOH)=4.76 log⁡2=0.30log⁡3=0.48\log 2=0.30\log 3=0.48log2=0.30log3=0.48
Numerical answer
View written solutionFree

Correct answer: 458

  1. Calculate initial moles

  • Moles of acetic acid: n(CH3COOH)=0.1×0.050=0.005 moln(\mathrm{CH_3COOH})=0.1\times 0.050=0.005\ \text{mol}n(CH3​COOH)=0.1×0.050=0.005 mol

  • Moles of NaOH: n(NaOH)=0.1×0.020=0.002 moln(\mathrm{NaOH})=0.1\times 0.020=0.002\ \text{mol}n(NaOH)=0.1×0.020=0.002 mol

  1. Neutralization reaction

CH3COOH+OH−→CH3COO−+H2O\mathrm{CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O}CH3​COOH+OH−→CH3​COO−+H2​O

Since NaOH is less, it will be completely consumed.

After reaction:

  • Acetic acid left: 0.005−0.002=0.003 mol0.005-0.002=0.003\ \text{mol}0.005−0.002=0.003 mol
  • Acetate formed: 0.002 mol0.002\ \text{mol}0.002 mol
  1. This is a buffer solution

Use Henderson–Hasselbalch equation: pH=pKa+log⁡[salt][acid]\mathrm{pH}=\mathrm{p}K_a+\log\frac{[\mathrm{salt}]}{[\mathrm{acid}]}pH=pKa​+log[acid][salt]​

Since both are in same final volume, we can use mole ratio: pH=4.76+log⁡0.0020.003\mathrm{pH}=4.76+\log\frac{0.002}{0.003}pH=4.76+log0.0030.002​

pH=4.76+log⁡23\mathrm{pH}=4.76+\log\frac{2}{3}pH=4.76+log32​

Now, log⁡23=log⁡2−log⁡3=0.30−0.48=−0.18\log\frac{2}{3}=\log 2-\log 3=0.30-0.48=-0.18log32​=log2−log3=0.30−0.48=−0.18

So, pH=4.76−0.18=4.58\mathrm{pH}=4.76-0.18=4.58pH=4.76−0.18=4.58

  1. Match with required format

The question says pH is ‾×10−2\underline{\hspace{1cm}}\times 10^{-2}​×10−2.

4.58=458×10−24.58=458\times 10^{-2}4.58=458×10−2

Hence the required nearest integer is: 458\boxed{458}458​

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