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Question #44105

Difficulty: 8/10 | 🟢 Free

The Question

Calculate the exact hydrogen ion concentration [H+][H^+] of a 1.0×10−61.0 \times 10^{-6} M aqueous solution of a generic amphiprotic salt \ce{NaHA} at 298 K. \ Given for the diprotic acid \ce{H2A}: Ka1=1.0×10−6K_{a1} = 1.0 \times 10^{-6}, Ka2=1.0×10−10K_{a2} = 1.0 \times 10^{-10}. \ Assume Kw=1.0×10−14K_w = 1.0 \times 10^{-14} at 298 K.

Options

  • A.

    1.00×10−81.00 \times 10^{-8} mol/L

  • B.

    7.23×10−87.23 \times 10^{-8} mol/L

  • C.

    7.11×10−87.11 \times 10^{-8} mol/L

  • D.

    1.00×10−71.00 \times 10^{-7} mol/L

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Amphiprotic Ion pH|pH of Extremely Dilute Solutions|Enthalpy Definition

Why This Question Is Tricky

Common Mistakes

This problem tests the limits of the standard amphiprotic pH formula pH=(pKa1+pKa2)/2pH = (pK_{a1} + pK_{a2})/2.

Students often blindly apply this formula, yielding 1.00×10−81.00 \times 10^{-8} M.

A more advanced student might recognize the dilution and use the exact proton balance equation, but mistakenly assume [\ceHA−]≈C[\ce{HA^-}] \approx C, yielding 7.11×10−87.11 \times 10^{-8} M.

The ultimate trap is that at this concentration and Ka1K_{a1} value, the protonation of \ceHA−\ce{HA^-} to \ceH2A\ce{H2A} is significant, meaning the mass balance must be solved exactly.

This leads to a cubic equation, revealing the true [H+][H^+] is 7.23×10−87.23 \times 10^{-8} M.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterEquilibrium - Class 11 Part 1
ConceptAmphiprotic Ion pH
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree