An electrochemical cell at 298 K is constructed as follows: Pt | H ₂ ( g , 1 bar ) | HA (0.1 M ) Ag ^+ (0.01 M ) | Ag where HA is a weak monoprotic acid. If the cell potential is 0.92 V , the p K_a of the weak acid HA is _________. (Given: E^ _ Ag ^+/ Ag = 0.80 V and 2.303RT F = 0.06 V )

Step-by-step solution

The cell reaction is: 1 2 H ₂( g ) + Ag ^+( aq ) H ^+( aq ) + Ag ( s ) Here, n = 1 . Applying the Nernst equation: E_ cell = E^ _ cell - 0.06 n [ H ^+] [ Ag ^+] Given E_ cell = 0.92 V and E^ _ cell = 0.80 V - 0 V = 0.80 V . Substituting the values: 0.92 = 0.80 - 0.06 1 [ H ^+] 0.01 0.12 = -0.06 (100 [ H ^+]) (100 [ H ^+]) = -2 100 [ H ^+] = 10⁻² [ H ^+] = 10⁻⁴ M For the weak acid HA , the dissociation constant K_a is given by: K_a = [ H ^+]^2 C K_a = (10⁻⁴)^2 0.1 = 10⁻⁸ 10⁻¹ = 10⁻⁷ Therefore, p K_a = - K_a = 7 . An