Calculate the percent ionization of a 0.125 M aqueous HCN solution. For HCN, Ka = 4.9 10–10.

Question

Calculate the percent ionization of a 0.125 M aqueous HCN solution.  For HCN, Ka = 4.9  10–10.

BlackTom AI · Accepted Answer

Reframe the problem: we need the percent ionization of a 0.125 M HCN solution given Ka = 4.9×10^−10.

Step 1: Set up the equilibrium. For a weak acid HA like HCN, HA ⇌ H+ + A−. Let x be the amount ionized: [H+] = [CN−] = x, and [HCN] ≈ C − x where C = 0.125 M.

Step 2: Write the Ka expression. Ka = [H+][A−] / [HA] = x·x / (C − x) = x^2 / (C − x).

Step 3: Decide on an approximation. Because Ka is very small and C is relatively large, we expect x ≪ C, so (C − x) ≈ C. This simplifies the calculation significantly and is a standard approach for weak acids.

Step 4: Compute x using the approximation. x ≈ sqrt(Ka × C) = sqrt((4.9×10^−10) × 0.125).

Step 5: Do the arithmetic. The product Ka × C is 4.9×10^−10 × 0.125 = 6.125×10^−11. The square root is x ≈ 7.83×10^−6 M.

Step 6: Convert to percent ionization. Percent ionization = (x / C) × 100% = (7.83×10^−6 / 0.125) × 100%. Compute the quotient: 7.83×10^−6 / 0.125 ≈ 6.264×10^−5. Multiply by 100%: ≈ 6.264×10^−3%, which rounds to about 0.0063%.

Step 7: Optional check. If one solves without approximation, x would satisfy x^2/(C − x) = Ka. With such a tiny Ka and small x, the approximation x ≈ sqrt(Ka C) is valid and the result remains effectively the same to several significant figures. If you kept the (C − x) term, you'd obtain a negligible difference given x ≪ C, confirming the reliability of the approximation.

Final note: The requested output structure lacks explicit answer choices, so the explanation focuses on the calculation steps, reasoning, and justification for using the standard approximation in this weak-acid scenario.

Chem 1C General Chemis... Self Assessment Quiz 13

Calculate the percent ionization of a 0.125 M aqueous HCN solution. For HCN, Ka = 4.9 10–10.

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