A galvanic cell is constructed using a standard copper electrode and a hydrogen electrode immersed in a 0.1 M solution of a weak monobasic acid HA . The cell representation is: Pt H ₂( g , 1 atm ) HA ( aq , 0.1 M ) Cu ²⁺( aq , 0.1 M ) Cu ( s ) If the measured cell potential is 0.49 V at 298 K , the p K_a of the weak acid HA is ________. (Nearest integer) (Given: E^ _ Cu ²⁺/ Cu = 0.34 V and 2.303RT F = 0.06 V )

Step-by-step solution

The cell reactions are: Anode: H ₂( g ) 2 H ^+( aq ) + 2 e ^- Cathode: Cu ²⁺( aq ) + 2 e ^- Cu ( s ) Overall: H ₂( g ) + Cu ²⁺( aq ) 2 H ^+( aq ) + Cu ( s ) The standard cell potential is: E^ _ cell = E^ _ cathode - E^ _ anode = 0.34 - 0 = 0.34 V Applying the Nernst equation: E_ cell = E^ _ cell - 0.06 2 [ H ^+]^2 P_ H ₂ [ Cu ²⁺] Substitute the given values ( E_ cell = 0.49 V , [ Cu ²⁺] = 0.1 M , P_ H ₂ = 1 atm ): 0.49 = 0.34 - 0.03 [ H ^+]^2 0.1 0.15 = -0.03 [ H ^+]^2 0.1 -5 = [ H ^+]^2 0.1 [ H ^+]^2 0.1 = 10⁻⁵ [