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Ionic Equilibrium question

2018 · 15 Apr · Shift 2 · Q19
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Ionic Equilibrium question

2018 · 15 Apr · Shift 2 · Q19

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
Following four solutions are prepared by mixing different volumes of NaOH and HCl of different concentrations, pH of which one of them will be equal to 1 ?
  1. A
    100 mL M10{M \over {10}}10M​ HCl + 100 mL M10{M \over {10}}10M​ NaOH
  2. B
    75 mL M5{M \over {5}}5M​ HCl + 25 mL M5{M \over {5}}5M​ NaOH
  3. C
    60 mL M10{M \over {10}}10M​ HCl + 40 mL M10{M \over {10}}10M​ NaOH
  4. D
    55 mL M10{M \over {10}}10M​ HCl + 45 mL M10{M \over {10}}10M​ NaOH
View written solutionFree

Correct answer: B

  1. Idea: Since HCl and NaOH are strong acid and strong base, they neutralize completely.

    HCl+NaOH→NaCl+H2O\text{HCl} + \text{NaOH} \to \text{NaCl} + \text{H}_2\text{O}HCl+NaOH→NaCl+H2​O

    After mixing, if acid is left in excess, then

    [H+]=moles of excess HCltotal volume in L[H^+] = \frac{\text{moles of excess HCl}}{\text{total volume in L}}[H+]=total volume in Lmoles of excess HCl​

    and

    pH=−log⁡[H+]\text{pH} = -\log [H^+]pH=−log[H+]

    For pH to be equal to 111,

    [H+]=10−1=0.1 M[H^+] = 10^{-1} = 0.1\,\text{M}[H+]=10−1=0.1M

  1. Check option A

    Given:

    • 100 mL100\,\text{mL}100mL of M10=0.1 M\dfrac{M}{10} = 0.1\,\text{M}10M​=0.1M HCl
    • 100 mL100\,\text{mL}100mL of 0.1 M0.1\,\text{M}0.1M NaOH

    Moles of HCl: 0.1×0.100=0.0100.1 \times 0.100 = 0.0100.1×0.100=0.010

    Moles of NaOH: 0.1×0.100=0.0100.1 \times 0.100 = 0.0100.1×0.100=0.010

    They completely neutralize each other. Final solution is neutral: pH=7\text{pH} = 7pH=7

    So, A is not correct.

  1. Check option B

    Given:

    • 75 mL75\,\text{mL}75mL of M5=0.2 M\dfrac{M}{5} = 0.2\,\text{M}5M​=0.2M HCl
    • 25 mL25\,\text{mL}25mL of 0.2 M0.2\,\text{M}0.2M NaOH

    Moles of HCl: 0.2×0.075=0.0150.2 \times 0.075 = 0.0150.2×0.075=0.015

    Moles of NaOH: 0.2×0.025=0.0050.2 \times 0.025 = 0.0050.2×0.025=0.005

    Excess HCl: 0.015−0.005=0.0100.015 - 0.005 = 0.0100.015−0.005=0.010

    Total volume: 75+25=100 mL=0.100 L75 + 25 = 100\,\text{mL} = 0.100\,\text{L}75+25=100mL=0.100L

    Therefore, [H+]=0.0100.100=0.1 M[H^+] = \frac{0.010}{0.100} = 0.1\,\text{M}[H+]=0.1000.010​=0.1M

    Hence, pH=−log⁡(0.1)=1\text{pH} = -\log(0.1) = 1pH=−log(0.1)=1

    So, B is correct.

  1. Check option C

    Given:

    • 60 mL60\,\text{mL}60mL of 0.1 M0.1\,\text{M}0.1M HCl
    • 40 mL40\,\text{mL}40mL of 0.1 M0.1\,\text{M}0.1M NaOH

    Moles of HCl: 0.1×0.060=0.0060.1 \times 0.060 = 0.0060.1×0.060=0.006

    Moles of NaOH: 0.1×0.040=0.0040.1 \times 0.040 = 0.0040.1×0.040=0.004

    Excess HCl: 0.006−0.004=0.0020.006 - 0.004 = 0.0020.006−0.004=0.002

    Total volume: 0.100 L0.100\,\text{L}0.100L

    [H+]=0.0020.100=0.02 M[H^+] = \frac{0.002}{0.100} = 0.02\,\text{M}[H+]=0.1000.002​=0.02M

    pH=−log⁡(0.02)≈1.70\text{pH} = -\log(0.02) \approx 1.70pH=−log(0.02)≈1.70

    So, C is not correct.

  1. Check option D

    Given:

    • 55 mL55\,\text{mL}55mL of 0.1 M0.1\,\text{M}0.1M HCl
    • 45 mL45\,\text{mL}45mL of 0.1 M0.1\,\text{M}0.1M NaOH

    Moles of HCl: 0.1×0.055=0.00550.1 \times 0.055 = 0.00550.1×0.055=0.0055

    Moles of NaOH: 0.1×0.045=0.00450.1 \times 0.045 = 0.00450.1×0.045=0.0045

    Excess HCl: 0.0055−0.0045=0.00100.0055 - 0.0045 = 0.00100.0055−0.0045=0.0010

    Total volume: 0.100 L0.100\,\text{L}0.100L

    [H+]=0.00100.100=0.01 M[H^+] = \frac{0.0010}{0.100} = 0.01\,\text{M}[H+]=0.1000.0010​=0.01M

    pH=−log⁡(0.01)=2\text{pH} = -\log(0.01) = 2pH=−log(0.01)=2

    So, D is not correct.

  1. Conclusion

    Only option B gives pH=1\text{pH} = 1pH=1

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