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Ionic Equilibrium question

2023 · 12 Apr · Shift 1 · Q21
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  5. /2023 · 12 Apr · Shift 1 · Q21

Ionic Equilibrium question

2023 · 12 Apr · Shift 1 · Q21

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
An analyst wants to convert 1 L HCl1 \mathrm{~L} \mathrm{~HCl}1 L HCl of pH=1\mathrm{pH}=1pH=1 to a solution of HCl\mathrm{HCl}HCl of pH 2\mathrm{pH} ~2pH 2. The volume of water needed to do this dilution is ‾mL\underline{\hspace{2cm}}\mathrm{mL}​mL. (Nearest integer)
Numerical answer
View written solutionFree

Correct answer: 9000

  1. Find initial concentration of HCl

Given initial pH=1\mathrm{pH} = 1pH=1.

For strong acid HCl, [H+]=10−pH=10−1=0.1 M[\mathrm{H}^+] = 10^{-\mathrm{pH}} = 10^{-1} = 0.1\,\text{M}[H+]=10−pH=10−1=0.1M

So the initial concentration of HCl is: C1=0.1 MC_1 = 0.1\,\text{M}C1​=0.1M

Initial volume: V1=1 LV_1 = 1\,\text{L}V1​=1L

  1. Find final concentration required

Target pH=2\mathrm{pH} = 2pH=2.

Thus, [H+]=10−2=0.01 M[\mathrm{H}^+] = 10^{-2} = 0.01\,\text{M}[H+]=10−2=0.01M

So final concentration should be: C2=0.01 MC_2 = 0.01\,\text{M}C2​=0.01M

  1. Use dilution formula

Since moles of HCl remain constant during dilution, C1V1=C2V2C_1V_1 = C_2V_2C1​V1​=C2​V2​

Substitute values: 0.1×1=0.01×V20.1 \times 1 = 0.01 \times V_20.1×1=0.01×V2​

V2=0.10.01=10 LV_2 = \frac{0.1}{0.01} = 10\,\text{L}V2​=0.010.1​=10L

  1. Calculate volume of water added

Initial volume is 1 L1\,\text{L}1L, final volume is 10 L10\,\text{L}10L.

Therefore, water added: 10−1=9 L10 - 1 = 9\,\text{L}10−1=9L

Convert to mL: 9 L=9000 mL9\,\text{L} = 9000\,\text{mL}9L=9000mL

  1. Final answer

The volume of water needed is: 9000 mL\boxed{9000\ \text{mL}}9000 mL​

  1. Comparison with stored answer

Stored correct answer = 900090009000.

My derived answer matches the stored answer.

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