Which of the following solutions gives the sharpest turning point in titration with $0\mathrm{.}1$ M HCl$.$

a$)0\mathrm{.}1$ M NaOCl Ka$($HOCl$)-3\mathrm{.}0$x$10^-8$ and $0\mathrm{.}1$M ethylamine solution Ka$($C$2$H$5$NH$3)=2\mathrm{.}31$x$10^-11$

b$)0\mathrm{.}1$ M NH$3$ Ka$($ NH$4=5\mathrm{.}70$x $10^-10$ and $0\mathrm{.}1$M sodiumphenolate $($Ka$($phenol$)=1\mathrm{.}0$x$10^-10)$