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Ionic Equilibrium question

2022 · 26 Jun · Shift 1 · Q14
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  5. /2022 · 26 Jun · Shift 1 · Q14

Ionic Equilibrium question

2022 · 26 Jun · Shift 1 · Q14

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
50 mL of 0.1 M CH3COOHCH_3COOHCH3​COOH is being titrated against 0.1 M NaOHNaOHNaOH. When 25 mL of NaOHNaOHNaOH has been added, the pH of the solution will be ‾×\underline{\hspace{2cm}}\times​× 10 −-− 2. (Nearest integer) (Given : pKa (CH3COOHCH_3COOHCH3​COOH) = 4.76) log 2 = 0.30 log 3 = 0.48 log 5 = 0.69 log 7 = 0.84 log 11 = 1.04
Numerical answer
View written solutionFree

Correct answer: 476

  1. Initial moles of acetic acid

Given:

  • Volume of CH3COOH=50 mL=0.050 LCH_3COOH = 50\,\text{mL} = 0.050\,\text{L}CH3​COOH=50mL=0.050L
  • Concentration =0.1 M= 0.1\,M=0.1M

So, n(CH3COOH)=0.1×0.050=0.005 moln(CH_3COOH)=0.1\times 0.050=0.005\,\text{mol}n(CH3​COOH)=0.1×0.050=0.005mol

  1. Moles of NaOH added

Given:

  • Volume of NaOH=25 mL=0.025 LNaOH = 25\,\text{mL} = 0.025\,\text{L}NaOH=25mL=0.025L
  • Concentration =0.1 M= 0.1\,M=0.1M

Thus, n(NaOH)=0.1×0.025=0.0025 moln(NaOH)=0.1\times 0.025=0.0025\,\text{mol}n(NaOH)=0.1×0.025=0.0025mol

  1. Neutralization reaction

CH3COOH+OH−→CH3COO−+H2OCH_3COOH + OH^- \rightarrow CH_3COO^- + H_2OCH3​COOH+OH−→CH3​COO−+H2​O

Initially:

  • CH3COOH=0.005CH_3COOH = 0.005CH3​COOH=0.005 mol
  • OH−=0.0025OH^- = 0.0025OH−=0.0025 mol

After reaction:

  • Acid left =0.005−0.0025=0.0025= 0.005-0.0025=0.0025=0.005−0.0025=0.0025 mol
  • Salt formed =0.0025= 0.0025=0.0025 mol

So, after adding 25 mL25\,\text{mL}25mL NaOH, we have equal moles of CH3COOHCH_3COOHCH3​COOH and CH3COO−CH_3COO^-CH3​COO−.

  1. Use Henderson–Hasselbalch equation

pH=pKa+log⁡[salt][acid]\text{pH} = \text{p}K_a + \log \frac{[salt]}{[acid]}pH=pKa​+log[acid][salt]​

Since moles of salt and acid are equal, [salt][acid]=1\frac{[salt]}{[acid]}=1[acid][salt]​=1

Therefore, pH=4.76+log⁡1=4.76\text{pH} = 4.76 + \log 1 = 4.76pH=4.76+log1=4.76

  1. Match with required form

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

Now, 4.76=476×10−24.76 = 476 \times 10^{-2}4.76=476×10−2

Hence, the required integer is 476\boxed{476}476​

  1. Comparison with stored answer

Stored correct answer = 476476476

This matches our derived answer.

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