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Ionic Equilibrium question

2025 · Shift 1 · Q9
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Ionic Equilibrium question

2025 · Shift 1 · Q9

JEE AdvancedChemistryIonic EquilibriumNumerical+4 / −1
At 25 °C, the concentration of H+H^+H+ ions in 1.00 × 10−3 M aqueous solution of a weak monobasic acid having acid dissociation constant (Ka) of 4.00 × 10−11 is X × 10−7 M. The value of X is ‾\underline{\hspace{2cm}}​. Use: Ionic product of water (Kw) = 1.00 × 10−14 at 25 °C
Numerical answer
View written solutionFree

Correct answer: 2.2TO2.3

Step-by-step Solution

  1. Identify the Equilibria: In an aqueous solution of a weak monobasic acid (HA), two equilibria exist simultaneously:

    • Dissociation of the weak acid: HA⇌H++A−HA \rightleftharpoons H^+ + A^-HA⇌H++A−
    • Autoionization of water: H2O⇌H++OH−H_2O \rightleftharpoons H^+ + OH^-H2​O⇌H++OH−

  2. Equilibrium Constants:

    • Acid dissociation constant: Ka=[H+][A−][HA]=4.00×10−11K_a = \frac{[H^+][A^-]}{[HA]} = 4.00 \times 10^{-11}Ka​=[HA][H+][A−]​=4.00×10−11
    • Ionic product of water: Kw=[H+][OH−]=1.00×10−14K_w = [H^+][OH^-] = 1.00 \times 10^{-14}Kw​=[H+][OH−]=1.00×10−14

  3. Check the Significance of Water Autoionization: Let's first calculate the [H+][H^+][H+] assuming it comes only from the acid dissociation, using the formula [H+]=KaC[H^+] = \sqrt{K_a C}[H+]=Ka​C​, where CCC is the initial concentration of the acid. C=1.00×10−3C = 1.00 \times 10^{-3}C=1.00×10−3 M [H+]acid≈(4.00×10−11)(1.00×10−3)=4.00×10−14=2.00×10−7[H^+]_{acid} \approx \sqrt{(4.00 \times 10^{-11})(1.00 \times 10^{-3})} = \sqrt{4.00 \times 10^{-14}} = 2.00 \times 10^{-7}[H+]acid​≈(4.00×10−11)(1.00×10−3)​=4.00×10−14​=2.00×10−7 M. This calculated hydrogen ion concentration (2.00×10−72.00 \times 10^{-7}2.00×10−7 M) is very close to the concentration of H+H^+H+ from pure water (1.00×10−71.00 \times 10^{-7}1.00×10−7 M). This indicates that the contribution of H+H^+H+ ions from the autoionization of water is significant and cannot be neglected.

  4. Set up the Full Equilibrium Calculation: We must consider all sources of ions. The principle of charge neutrality states that the total positive charge concentration must equal the total negative charge concentration in the solution. [H+]=[A−]+[OH−][H^+] = [A^-] + [OH^-][H+]=[A−]+[OH−]

  5. Express concentrations in terms of [H+][H^+][H+]:

    • From the KwK_wKw​ expression: [OH−]=Kw[H+][OH^-] = \frac{K_w}{[H^+]}[OH−]=[H+]Kw​​
    • From the KaK_aKa​ expression: [A−]=Ka[HA][H+][A^-] = \frac{K_a[HA]}{[H^+]}[A−]=[H+]Ka​[HA]​ Since the acid is very weak (Ka=4.00×10−11K_a = 4.00 \times 10^{-11}Ka​=4.00×10−11), its dissociation is very small. Therefore, the equilibrium concentration of the undissociated acid, [HA][HA][HA], is approximately equal to its initial concentration, CCC. [HA]≈C=1.00×10−3[HA] \approx C = 1.00 \times 10^{-3}[HA]≈C=1.00×10−3 M. So, we can approximate [A−][A^-][A−] as: [A−]≈KaC[H+][A^-] \approx \frac{K_a C}{[H^+]}[A−]≈[H+]Ka​C​

  6. Solve for [H+][H^+][H+]: Substitute the expressions for [A−][A^-][A−] and [OH−][OH^-][OH−] into the charge neutrality equation: [H+]=KaC[H+]+Kw[H+][H^+] = \frac{K_a C}{[H^+]} + \frac{K_w}{[H^+]}[H+]=[H+]Ka​C​+[H+]Kw​​ Multiply the entire equation by [H+][H^+][H+]: [H+]2=KaC+Kw[H^+]^2 = K_a C + K_w[H+]2=Ka​C+Kw​ Now, we can solve for [H+][H^+][H+]: [H+]=KaC+Kw[H^+] = \sqrt{K_a C + K_w}[H+]=Ka​C+Kw​​

  7. Calculate the Numerical Value: Substitute the given values into the equation: C=1.00×10−3C = 1.00 \times 10^{-3}C=1.00×10−3 M Ka=4.00×10−11K_a = 4.00 \times 10^{-11}Ka​=4.00×10−11 Kw=1.00×10−14K_w = 1.00 \times 10^{-14}Kw​=1.00×10−14 [H+]=(4.00×10−11)(1.00×10−3)+1.00×10−14[H^+] = \sqrt{(4.00 \times 10^{-11})(1.00 \times 10^{-3}) + 1.00 \times 10^{-14}}[H+]=(4.00×10−11)(1.00×10−3)+1.00×10−14​ [H+]=4.00×10−14+1.00×10−14[H^+] = \sqrt{4.00 \times 10^{-14} + 1.00 \times 10^{-14}}[H+]=4.00×10−14+1.00×10−14​ [H+]=5.00×10−14[H^+] = \sqrt{5.00 \times 10^{-14}}[H+]=5.00×10−14​ [H+]=5×10−7 M[H^+] = \sqrt{5} \times 10^{-7} \text{ M}[H+]=5​×10−7 M The value of 5\sqrt{5}5​ is approximately 2.236. [H+]≈2.236×10−7 M[H^+] \approx 2.236 \times 10^{-7} \text{ M}[H+]≈2.236×10−7 M

  8. Determine the value of X: The problem states that the concentration of H+H^+H+ ions is X×10−7X \times 10^{-7}X×10−7 M. Comparing our calculated value with the given format: X×10−7=2.236×10−7X \times 10^{-7} = 2.236 \times 10^{-7}X×10−7=2.236×10−7 Therefore, the value of X is 2.236.

  9. Final Answer Comparison: The calculated value X=2.236X = 2.236X=2.236 falls within the given answer range of 2.2 to 2.3.

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