Find the pH of a 0.175 M aqueous NaCN solution. For HCN, Ka = 4.9 10–10.

Question

Find the pH of a 0.175 M aqueous NaCN solution.  For HCN, Ka = 4.9  10–10.

BlackTom AI · Accepted Answer

The problem asks for the pH of a 0.175 M solution of NaCN, given Ka for HCN = 4.9e-10. CN- is the conjugate base of a weak acid HCN, so the CN- in solution hydrolyzes: CN- + H2O ⇌ HCN + OH-. The relevant equilibrium constant is Kb for CN-, which is Kw/Ka(HCN) = 1.0e-14 / 4.9e-10 ≈ 2.04e-5.

With a fairly concentrated solution, we can use the common approximation for a weak base: [OH-] ≈ sqrt(Kb * Cb), where Cb is the base concentration, 0.175 M. Compute: Kb * Cb = 2.04e-5 × 0.175 ≈ 3.57e-6. Square root gives [OH-] ≈ 1.89e-3 M.

Then pOH = -log10([OH-]) ≈ 2.72, and pH = 14.00 − pOH ≈ 11.28.

This uses the assumption that x is small relative to Cb and that water autoionization is negligible in comparison; also note that the salt NaCN fully dissociates, so initial CN- concentration equals 0.175 M.

If the options included a different approach or used a more exact expression, similar steps would be followed, but the result would be very close to 11.28.

Find the pH of a 0.175 M aqueous NaCN solution. For HCN, Ka = 4.9 10–10.单项选择题

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