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Ionic Equilibrium question

2022 · 28 Jul · Shift 1 · Q17
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  5. /2022 · 28 Jul · Shift 1 · Q17

Ionic Equilibrium question

2022 · 28 Jul · Shift 1 · Q17

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
Ka\mathrm{K}_{\mathrm{a}}Ka​ for butyric acid (C3H7COOH)\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{COOH}\right)(C3​H7​COOH) is 2×10−52 \times 10^{-5}2×10−5. The pH\mathrm{pH}pH of 0.2 M0.2 \,\mathrm{M}0.2M solution of butyric acid is ‾\underline{\hspace{2cm}}​×10−1\times 10^{-1}×10−1. (Nearest integer) [Given log⁡2=0.30\log 2=0.30log2=0.30]
Numerical answer
View written solutionFree

Correct answer: 27

  1. Given data

    • Acid: butyric acid, a weak monoprotic acid
    • Ka=2×10−5K_a = 2 \times 10^{-5}Ka​=2×10−5
    • Initial concentration, C=0.2 MC = 0.2\,\text{M}C=0.2M

  2. Use weak acid approximation

    For a weak acid HAHAHA of concentration CCC: [H+]≈KaC[H^+] \approx \sqrt{K_a C}[H+]≈Ka​C​

    Substituting: [H+]=(2×10−5)(0.2)[H^+] = \sqrt{(2 \times 10^{-5})(0.2)}[H+]=(2×10−5)(0.2)​

    Since 0.2=2×10−10.2 = 2 \times 10^{-1}0.2=2×10−1, [H+]=2×10−5×2×10−1[H^+] = \sqrt{2 \times 10^{-5} \times 2 \times 10^{-1}}[H+]=2×10−5×2×10−1​ =4×10−6= \sqrt{4 \times 10^{-6}}=4×10−6​ =2×10−3= 2 \times 10^{-3}=2×10−3

  3. Find pH

    pH=−log⁡(2×10−3)\text{pH} = -\log(2 \times 10^{-3})pH=−log(2×10−3) =−[log⁡2+log⁡10−3]= -[\log 2 + \log 10^{-3}]=−[log2+log10−3] =−(0.30−3)= -(0.30 - 3)=−(0.30−3) =2.70= 2.70=2.70

  4. Match with the format in the question

    The question writes pH as: ‾×10−1\underline{\hspace{2cm}} \times 10^{-1}​×10−1

    Since 2.70=27×10−12.70 = 27 \times 10^{-1}2.70=27×10−1

    the required integer is: 27\boxed{27}27​

  5. Comparison with stored answer

    Derived answer = 272727

    Stored correct answer = 272727

    They match.

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