# How do you calculate the pH at the equivalence point for the titration of 0.140 M methylamine (CH_3NH_2) with 0.140 M HCI? The K_b of methylamine is 5.0x10^-4?

at the equivalent point you have $C {H}_{3} N {H}_{3} C l$, and having unitet the two solutions you have a concentration of 0,07 M, The pH is given by
$\left[O {H}^{-}\right] = \sqrt{C s \frac{K w}{K a}} = \sqrt{0 , 07 \times {10}^{- 14} / \left(5 \times {10}^{- 5}\right)} = 3 , 7 \times {10}^{- 6}$