A hydrogen electrode ( $p_{H X_{2}} = 1 bar$ ) dipped in $0.10 M$ solution of a weak monobasic acid HA is coupled with a standard hydrogen electrode. The EMF is $0.1477 V$ at 298 K. Use $K_{a} = [H X^{+}]^{2} / c$ (neglect dissociation relative to $c$ ) and $2.303 RT / F = 0.0591 V$ . The $p K_{a}$ of HA is (one decimal).

Numerical answer type, tolerance 0.05.

Answer: 4

The HA solution has lower

[H X^{+}]

, so its electrode is the anode and

E_{cell} = 0 - (- 0.0591 pH) = 0.0591 pH

pH = 0.1477/0.0591 = 2.50

, so

[H X^{+}] = 1 0^{- 2.50} = 3.16 \times 1 0^{- 3} M

K_{a} = (3.16 \times 1 0^{- 3})^{2} /0.10 = 1.0 \times 1 0^{- 4}

, i.e.

p K_{a} = 2 pH + lo g c = 5.00 - 1 = 4.0