Given the following molar conductivities at 25°C; HCl, 426${\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$; NaCl, 126 ${\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$; NaC (sodium crotonate), 83${\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$, x × 10^–5 is the ionization constant of crotonic acid, if the conductivity of a 0.001 M crotonic acid solution is 3.83 x10^-5 ${\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{-1}\cdot \mathrm{x}=$