Question

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`?

Answer 1

pH = 8,57

Explanation:

at the equivalent point you have `CH_3NH_3Cl`, and having unitet the two solutions you have a concentration of 0,07 M, The pH is given by
`[OH^-]=sqrt(Cs(Kw)/(Ka)) =sqrt (0,07 xx 10^(-14)/(5 xx 10^(-5))) = 3,7 xx 10^(-6)`
pOH= 5.43; pH = 8,57